Algebra Examples

$\begin{array}{c} x & y \\ 3 & 1 \\ 7 & 2 \\ 11 & 3 \\ 18 & 5 \end{array}$

Step 1

Check if the function rule is linear.

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Step 1.1

To find if the table follows a function rule, check to see if the values follow the linear form $y = a x + b$ .

$y = a x + b$

Step 1.2

Build a set of equations from the table such that $y = a x + b$ .

$1 = a (3) + b 2 = a (7) + b 3 = a (11) + b 5 = a (18) + b$

Step 1.3

Calculate the values of $a$ and $b$ .

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Step 1.3.1

Solve for $b$ in $1 = a (3) + b$ .

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Step 1.3.1.1

Rewrite the equation as $a (3) + b = 1$ .

$a (3) + b = 1$

$2 = a (7) + b$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.1.2

Move $3$ to the left of $a$ .

$3 a + b = 1$

$2 = a (7) + b$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.1.3

Subtract $3 a$ from both sides of the equation.

$b = 1 - 3 a$

$2 = a (7) + b$

$3 = a (11) + b$

$5 = a (18) + b$

$b = 1 - 3 a$

$2 = a (7) + b$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.2

Replace all occurrences of $b$ with $1 - 3 a$ in each equation.

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Step 1.3.2.1

Replace all occurrences of $b$ in $2 = a (7) + b$ with $1 - 3 a$ .

$2 = a (7) + 1 - 3 a$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.2.2

Simplify $2 = a (7) + 1 - 3 a$ .

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Step 1.3.2.2.1

Simplify the left side.

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Step 1.3.2.2.1.1

Remove parentheses.

$2 = a (7) + 1 - 3 a$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

$2 = a (7) + 1 - 3 a$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.2.2.2

Simplify the right side.

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Step 1.3.2.2.2.1

Simplify $a (7) + 1 - 3 a$ .

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Step 1.3.2.2.2.1.1

Move $7$ to the left of $a$ .

$2 = 7 a + 1 - 3 a$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.2.2.2.1.2

Subtract $3 a$ from $7 a$ .

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = a (11) + b$

$5 = a (18) + b$

Step 1.3.2.3

Replace all occurrences of $b$ in $3 = a (11) + b$ with $1 - 3 a$ .

$3 = a (11) + 1 - 3 a$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

Step 1.3.2.4

Simplify $3 = a (11) + 1 - 3 a$ .

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Step 1.3.2.4.1

Simplify the left side.

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Step 1.3.2.4.1.1

Remove parentheses.

$3 = a (11) + 1 - 3 a$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

$3 = a (11) + 1 - 3 a$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

Step 1.3.2.4.2

Simplify the right side.

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Step 1.3.2.4.2.1

Simplify $a (11) + 1 - 3 a$ .

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Step 1.3.2.4.2.1.1

Move $11$ to the left of $a$ .

$3 = 11 a + 1 - 3 a$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

Step 1.3.2.4.2.1.2

Subtract $3 a$ from $11 a$ .

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + b$

Step 1.3.2.5

Replace all occurrences of $b$ in $5 = a (18) + b$ with $1 - 3 a$ .

$5 = a (18) + 1 - 3 a$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.2.6

Simplify $5 = a (18) + 1 - 3 a$ .

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Step 1.3.2.6.1

Simplify the left side.

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Step 1.3.2.6.1.1

Remove parentheses.

$5 = a (18) + 1 - 3 a$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = a (18) + 1 - 3 a$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.2.6.2

Simplify the right side.

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Step 1.3.2.6.2.1

Simplify $a (18) + 1 - 3 a$ .

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Step 1.3.2.6.2.1.1

Move $18$ to the left of $a$ .

$5 = 18 a + 1 - 3 a$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.2.6.2.1.2

Subtract $3 a$ from $18 a$ .

$5 = 15 a + 1$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = 15 a + 1$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = 15 a + 1$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = 15 a + 1$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$5 = 15 a + 1$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.3

Solve for $a$ in $5 = 15 a + 1$ .

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Step 1.3.3.1

Rewrite the equation as $15 a + 1 = 5$ .

$15 a + 1 = 5$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 1.3.3.2.1

Subtract $1$ from both sides of the equation.

$15 a = 5 - 1$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.3.2.2

Subtract $1$ from $5$ .

$15 a = 4$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$15 a = 4$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.3.3

Divide each term in $15 a = 4$ by $15$ and simplify.

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Step 1.3.3.3.1

Divide each term in $15 a = 4$ by $15$ .

$\frac{15 a}{15} = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.3.3.2

Simplify the left side.

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Step 1.3.3.3.2.1

Cancel the common factor of $15$ .

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Step 1.3.3.3.2.1.1

Cancel the common factor.

$\frac{15 a}{15} = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$a = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$a = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$a = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

$a = \frac{4}{15}$

$3 = 8 a + 1$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4

Replace all occurrences of $a$ with $\frac{4}{15}$ in each equation.

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Step 1.3.4.1

Replace all occurrences of $a$ in $3 = 8 a + 1$ with $\frac{4}{15}$ .

$3 = 8 (\frac{4}{15}) + 1$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4.2

Simplify the right side.

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Step 1.3.4.2.1

Simplify $8 (\frac{4}{15}) + 1$ .

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Step 1.3.4.2.1.1

Multiply $8 (\frac{4}{15})$ .

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Step 1.3.4.2.1.1.1

Combine $8$ and $\frac{4}{15}$ .

$3 = \frac{8 \cdot 4}{15} + 1$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4.2.1.1.2

Multiply $8$ by $4$ .

$3 = \frac{32}{15} + 1$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = \frac{32}{15} + 1$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4.2.1.2

Write $1$ as a fraction with a common denominator.

$3 = \frac{32}{15} + \frac{15}{15}$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4.2.1.3

Combine the numerators over the common denominator.

$3 = \frac{32 + 15}{15}$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4.2.1.4

Add $32$ and $15$ .

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$2 = 4 a + 1$

$b = 1 - 3 a$

Step 1.3.4.3

Replace all occurrences of $a$ in $2 = 4 a + 1$ with $\frac{4}{15}$ .

$2 = 4 (\frac{4}{15}) + 1$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

Step 1.3.4.4

Simplify the right side.

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Step 1.3.4.4.1

Simplify $4 (\frac{4}{15}) + 1$ .

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Step 1.3.4.4.1.1

Multiply $4 (\frac{4}{15})$ .

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Step 1.3.4.4.1.1.1

Combine $4$ and $\frac{4}{15}$ .

$2 = \frac{4 \cdot 4}{15} + 1$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

Step 1.3.4.4.1.1.2

Multiply $4$ by $4$ .

$2 = \frac{16}{15} + 1$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

$2 = \frac{16}{15} + 1$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

Step 1.3.4.4.1.2

Write $1$ as a fraction with a common denominator.

$2 = \frac{16}{15} + \frac{15}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

Step 1.3.4.4.1.3

Combine the numerators over the common denominator.

$2 = \frac{16 + 15}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

Step 1.3.4.4.1.4

Add $16$ and $15$ .

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 3 a$

Step 1.3.4.5

Replace all occurrences of $a$ in $b = 1 - 3 a$ with $\frac{4}{15}$ .

$b = 1 - 3 (\frac{4}{15})$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6

Simplify the right side.

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Step 1.3.4.6.1

Simplify $1 - 3 (\frac{4}{15})$ .

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Step 1.3.4.6.1.1

Simplify each term.

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Step 1.3.4.6.1.1.1

Cancel the common factor of $3$ .

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Step 1.3.4.6.1.1.1.1

Factor $3$ out of $- 3$ .

$b = 1 + 3 (- 1) (\frac{4}{15})$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.1.1.2

Factor $3$ out of $15$ .

$b = 1 + 3 \cdot (- 1 \frac{4}{3 \cdot 5})$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.1.1.3

Cancel the common factor.

$b = 1 + 3 \cdot (- 1 \frac{4}{3 \cdot 5})$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.1.1.4

Rewrite the expression.

$b = 1 - 1 (\frac{4}{5})$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - 1 (\frac{4}{5})$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.1.2

Rewrite $- 1 (\frac{4}{5})$ as $- (\frac{4}{5})$ .

$b = 1 - \frac{4}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = 1 - \frac{4}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.2

Simplify the expression.

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Step 1.3.4.6.1.2.1

Write $1$ as a fraction with a common denominator.

$b = \frac{5}{5} - \frac{4}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.2.2

Combine the numerators over the common denominator.

$b = \frac{5 - 4}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.4.6.1.2.3

Subtract $4$ from $5$ .

$b = \frac{1}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = \frac{1}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = \frac{1}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = \frac{1}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

$b = \frac{1}{5}$

$2 = \frac{31}{15}$

$3 = \frac{47}{15}$

$a = \frac{4}{15}$

Step 1.3.5

Since $2 = \frac{31}{15}$ is not true, there is no solution.

No solution

Step 1.4

Since $y \neq y$ for the corresponding $x$ values, the function is not linear.

The function is not linear

Step 2

Check if the function rule is quadratic.

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Step 2.1

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{2} + b x + c$ .

$y = a x^{2} + b x + c$

Step 2.2

Build a set of $3$ equations from the table such that $y = a x^{2} + b x + c$ .

Step 2.3

Calculate the values of $a$ , $b$ , and $c$ .

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Step 2.3.1

Solve for $c$ in $1 = a \cdot 3^{2} + b (3) + c$ .

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Step 2.3.1.1

Rewrite the equation as $a \cdot 3^{2} + b (3) + c = 1$ .

$a \cdot 3^{2} + b (3) + c = 1$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.1.2

Simplify each term.

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Step 2.3.1.2.1

Raise $3$ to the power of $2$ .

$a \cdot 9 + b (3) + c = 1$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.1.2.2

Move $9$ to the left of $a$ .

$9 \cdot a + b (3) + c = 1$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.1.2.3

Move $3$ to the left of $b$ .

$9 a + 3 b + c = 1$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$9 a + 3 b + c = 1$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.1.3

Move all terms not containing $c$ to the right side of the equation.

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Step 2.3.1.3.1

Subtract $9 a$ from both sides of the equation.

$3 b + c = 1 - 9 a$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.1.3.2

Subtract $3 b$ from both sides of the equation.

$c = 1 - 9 a - 3 b$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$c = 1 - 9 a - 3 b$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$c = 1 - 9 a - 3 b$

$2 = a \cdot 7^{2} + b (7) + c$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2

Replace all occurrences of $c$ with $1 - 9 a - 3 b$ in each equation.

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Step 2.3.2.1

Replace all occurrences of $c$ in $2 = a \cdot 7^{2} + b (7) + c$ with $1 - 9 a - 3 b$ .

$2 = a \cdot 7^{2} + b (7) + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.2

Simplify $2 = a \cdot 7^{2} + b (7) + 1 - 9 a - 3 b$ .

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Step 2.3.2.2.1

Simplify the left side.

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Step 2.3.2.2.1.1

Remove parentheses.

$2 = a \cdot 7^{2} + b (7) + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$2 = a \cdot 7^{2} + b (7) + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.2.2

Simplify the right side.

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Step 2.3.2.2.2.1

Simplify $a \cdot 7^{2} + b (7) + 1 - 9 a - 3 b$ .

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Step 2.3.2.2.2.1.1

Simplify each term.

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Step 2.3.2.2.2.1.1.1

Raise $7$ to the power of $2$ .

$2 = a \cdot 49 + b (7) + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.2.2.1.1.2

Move $49$ to the left of $a$ .

$2 = 49 \cdot a + b (7) + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.2.2.1.1.3

Move $7$ to the left of $b$ .

$2 = 49 a + 7 b + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$2 = 49 a + 7 b + 1 - 9 a - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.2.2.1.2

Simplify by adding terms.

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Step 2.3.2.2.2.1.2.1

Subtract $9 a$ from $49 a$ .

$2 = 40 a + 7 b + 1 - 3 b$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.2.2.1.2.2

Subtract $3 b$ from $7 b$ .

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = a \cdot 11^{2} + b (11) + c$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.3

Replace all occurrences of $c$ in $3 = a \cdot 11^{2} + b (11) + c$ with $1 - 9 a - 3 b$ .

$3 = a \cdot 11^{2} + b (11) + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.4

Simplify $3 = a \cdot 11^{2} + b (11) + 1 - 9 a - 3 b$ .

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Step 2.3.2.4.1

Simplify the left side.

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Step 2.3.2.4.1.1

Remove parentheses.

$3 = a \cdot 11^{2} + b (11) + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

$3 = a \cdot 11^{2} + b (11) + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.4.2

Simplify the right side.

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Step 2.3.2.4.2.1

Simplify $a \cdot 11^{2} + b (11) + 1 - 9 a - 3 b$ .

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Step 2.3.2.4.2.1.1

Simplify each term.

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Step 2.3.2.4.2.1.1.1

Raise $11$ to the power of $2$ .

$3 = a \cdot 121 + b (11) + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.4.2.1.1.2

Move $121$ to the left of $a$ .

$3 = 121 \cdot a + b (11) + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.4.2.1.1.3

Move $11$ to the left of $b$ .

$3 = 121 a + 11 b + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

$3 = 121 a + 11 b + 1 - 9 a - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.4.2.1.2

Simplify by adding terms.

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Step 2.3.2.4.2.1.2.1

Subtract $9 a$ from $121 a$ .

$3 = 112 a + 11 b + 1 - 3 b$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.4.2.1.2.2

Subtract $3 b$ from $11 b$ .

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + c$

Step 2.3.2.5

Replace all occurrences of $c$ in $5 = a \cdot 18^{2} + b (18) + c$ with $1 - 9 a - 3 b$ .

$5 = a \cdot 18^{2} + b (18) + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.2.6

Simplify $5 = a \cdot 18^{2} + b (18) + 1 - 9 a - 3 b$ .

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Step 2.3.2.6.1

Simplify the left side.

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Step 2.3.2.6.1.1

Remove parentheses.

$5 = a \cdot 18^{2} + b (18) + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = a \cdot 18^{2} + b (18) + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.2.6.2

Simplify the right side.

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Step 2.3.2.6.2.1

Simplify $a \cdot 18^{2} + b (18) + 1 - 9 a - 3 b$ .

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Step 2.3.2.6.2.1.1

Simplify each term.

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Step 2.3.2.6.2.1.1.1

Raise $18$ to the power of $2$ .

$5 = a \cdot 324 + b (18) + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.2.6.2.1.1.2

Move $324$ to the left of $a$ .

$5 = 324 \cdot a + b (18) + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.2.6.2.1.1.3

Move $18$ to the left of $b$ .

$5 = 324 a + 18 b + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = 324 a + 18 b + 1 - 9 a - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.2.6.2.1.2

Simplify by adding terms.

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Step 2.3.2.6.2.1.2.1

Subtract $9 a$ from $324 a$ .

$5 = 315 a + 18 b + 1 - 3 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.2.6.2.1.2.2

Subtract $3 b$ from $18 b$ .

$5 = 315 a + 15 b + 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = 315 a + 15 b + 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = 315 a + 15 b + 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = 315 a + 15 b + 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = 315 a + 15 b + 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$5 = 315 a + 15 b + 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3

Solve for $a$ in $5 = 315 a + 15 b + 1$ .

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Step 2.3.3.1

Rewrite the equation as $315 a + 15 b + 1 = 5$ .

$315 a + 15 b + 1 = 5$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 2.3.3.2.1

Subtract $15 b$ from both sides of the equation.

$315 a + 1 = 5 - 15 b$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.2.2

Subtract $1$ from both sides of the equation.

$315 a = 5 - 15 b - 1$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.2.3

Subtract $1$ from $5$ .

$315 a = - 15 b + 4$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$315 a = - 15 b + 4$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3

Divide each term in $315 a = - 15 b + 4$ by $315$ and simplify.

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Step 2.3.3.3.1

Divide each term in $315 a = - 15 b + 4$ by $315$ .

$\frac{315 a}{315} = \frac{- 15 b}{315} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.2

Simplify the left side.

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Step 2.3.3.3.2.1

Cancel the common factor of $315$ .

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Step 2.3.3.3.2.1.1

Cancel the common factor.

$\frac{315 a}{315} = \frac{- 15 b}{315} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 15 b}{315} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = \frac{- 15 b}{315} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = \frac{- 15 b}{315} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.3

Simplify the right side.

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Step 2.3.3.3.3.1

Simplify each term.

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Step 2.3.3.3.3.1.1

Cancel the common factor of $- 15$ and $315$ .

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Step 2.3.3.3.3.1.1.1

Factor $15$ out of $- 15 b$ .

$a = \frac{15 (- b)}{315} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.3.1.1.2

Cancel the common factors.

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Step 2.3.3.3.3.1.1.2.1

Factor $15$ out of $315$ .

$a = \frac{15 (- b)}{15 (21)} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.3.1.1.2.2

Cancel the common factor.

$a = \frac{15 (- b)}{15 \cdot 21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.3.1.1.2.3

Rewrite the expression.

$a = \frac{- b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = \frac{- b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = \frac{- b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.3.3.3.1.2

Move the negative in front of the fraction.

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = 112 a + 8 b + 1$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4

Replace all occurrences of $a$ with $- \frac{b}{21} + \frac{4}{315}$ in each equation.

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Step 2.3.4.1

Replace all occurrences of $a$ in $3 = 112 a + 8 b + 1$ with $- \frac{b}{21} + \frac{4}{315}$ .

$3 = 112 (- \frac{b}{21} + \frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2

Simplify the right side.

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Step 2.3.4.2.1

Simplify $112 (- \frac{b}{21} + \frac{4}{315}) + 8 b + 1$ .

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Step 2.3.4.2.1.1

Simplify each term.

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Step 2.3.4.2.1.1.1

Apply the distributive property.

$3 = 112 (- \frac{b}{21}) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.2

Cancel the common factor of $7$ .

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Step 2.3.4.2.1.1.2.1

Move the leading negative in $- \frac{b}{21}$ into the numerator.

$3 = 112 (\frac{- b}{21}) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.2.2

Factor $7$ out of $112$ .

$3 = 7 (16) (\frac{- b}{21}) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.2.3

Factor $7$ out of $21$ .

$3 = 7 \cdot (16 (\frac{- b}{7 \cdot 3})) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.2.4

Cancel the common factor.

$3 = 7 \cdot (16 (\frac{- b}{7 \cdot 3})) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.2.5

Rewrite the expression.

$3 = 16 (\frac{- b}{3}) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = 16 (\frac{- b}{3}) + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.3

Combine $16$ and $\frac{- b}{3}$ .

$3 = \frac{16 (- b)}{3} + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.4

Multiply $- 1$ by $16$ .

$3 = \frac{- 16 b}{3} + 112 (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.5

Cancel the common factor of $7$ .

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Step 2.3.4.2.1.1.5.1

Factor $7$ out of $112$ .

$3 = \frac{- 16 b}{3} + 7 (16) (\frac{4}{315}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.5.2

Factor $7$ out of $315$ .

$3 = \frac{- 16 b}{3} + 7 \cdot (16 (\frac{4}{7 \cdot 45})) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.5.3

Cancel the common factor.

$3 = \frac{- 16 b}{3} + 7 \cdot (16 (\frac{4}{7 \cdot 45})) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.5.4

Rewrite the expression.

$3 = \frac{- 16 b}{3} + 16 (\frac{4}{45}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = \frac{- 16 b}{3} + 16 (\frac{4}{45}) + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.6

Combine $16$ and $\frac{4}{45}$ .

$3 = \frac{- 16 b}{3} + \frac{16 \cdot 4}{45} + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.7

Multiply $16$ by $4$ .

$3 = \frac{- 16 b}{3} + \frac{64}{45} + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.1.8

Move the negative in front of the fraction.

$3 = - \frac{16 b}{3} + \frac{64}{45} + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = - \frac{16 b}{3} + \frac{64}{45} + 8 b + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.2

To write $8 b$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$3 = - \frac{16 b}{3} + 8 b \cdot \frac{3}{3} + \frac{64}{45} + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.3

Combine $8 b$ and $\frac{3}{3}$ .

$3 = - \frac{16 b}{3} + \frac{8 b \cdot 3}{3} + \frac{64}{45} + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.4

Combine the numerators over the common denominator.

$3 = \frac{- 16 b + 8 b \cdot 3}{3} + \frac{64}{45} + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.5

Find the common denominator.

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Step 2.3.4.2.1.5.1

Multiply $\frac{- 16 b + 8 b \cdot 3}{3}$ by $\frac{15}{15}$ .

$3 = \frac{- 16 b + 8 b \cdot 3}{3} \cdot \frac{15}{15} + \frac{64}{45} + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.5.2

Multiply $\frac{- 16 b + 8 b \cdot 3}{3}$ by $\frac{15}{15}$ .

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15}{3 \cdot 15} + \frac{64}{45} + 1$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.5.3

Write $1$ as a fraction with denominator $1$ .

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15}{3 \cdot 15} + \frac{64}{45} + \frac{1}{1}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.5.4

Multiply $\frac{1}{1}$ by $\frac{45}{45}$ .

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15}{3 \cdot 15} + \frac{64}{45} + \frac{1}{1} \cdot \frac{45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.5.5

Multiply $\frac{1}{1}$ by $\frac{45}{45}$ .

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15}{3 \cdot 15} + \frac{64}{45} + \frac{45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.5.6

Multiply $3$ by $15$ .

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15}{45} + \frac{64}{45} + \frac{45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15}{45} + \frac{64}{45} + \frac{45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.6

Combine the numerators over the common denominator.

$3 = \frac{(- 16 b + 8 b \cdot 3) \cdot 15 + 64 + 45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.7

Simplify each term.

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Step 2.3.4.2.1.7.1

Multiply $3$ by $8$ .

$3 = \frac{(- 16 b + 24 b) \cdot 15 + 64 + 45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.7.2

Add $- 16 b$ and $24 b$ .

$3 = \frac{8 b \cdot 15 + 64 + 45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.7.3

Multiply $15$ by $8$ .

$3 = \frac{120 b + 64 + 45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = \frac{120 b + 64 + 45}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.2.1.8

Add $64$ and $45$ .

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$2 = 40 a + 4 b + 1$

$c = 1 - 9 a - 3 b$

Step 2.3.4.3

Replace all occurrences of $a$ in $2 = 40 a + 4 b + 1$ with $- \frac{b}{21} + \frac{4}{315}$ .

$2 = 40 (- \frac{b}{21} + \frac{4}{315}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4

Simplify the right side.

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Step 2.3.4.4.1

Simplify $40 (- \frac{b}{21} + \frac{4}{315}) + 4 b + 1$ .

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Step 2.3.4.4.1.1

Simplify each term.

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Step 2.3.4.4.1.1.1

Apply the distributive property.

$2 = 40 (- \frac{b}{21}) + 40 (\frac{4}{315}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.2

Multiply $40 (- \frac{b}{21})$ .

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Step 2.3.4.4.1.1.2.1

Multiply $- 1$ by $40$ .

$2 = - 40 \frac{b}{21} + 40 (\frac{4}{315}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.2.2

Combine $- 40$ and $\frac{b}{21}$ .

$2 = \frac{- 40 b}{21} + 40 (\frac{4}{315}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = \frac{- 40 b}{21} + 40 (\frac{4}{315}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.3

Cancel the common factor of $5$ .

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Step 2.3.4.4.1.1.3.1

Factor $5$ out of $40$ .

$2 = \frac{- 40 b}{21} + 5 (8) (\frac{4}{315}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.3.2

Factor $5$ out of $315$ .

$2 = \frac{- 40 b}{21} + 5 \cdot (8 (\frac{4}{5 \cdot 63})) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.3.3

Cancel the common factor.

$2 = \frac{- 40 b}{21} + 5 \cdot (8 (\frac{4}{5 \cdot 63})) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.3.4

Rewrite the expression.

$2 = \frac{- 40 b}{21} + 8 (\frac{4}{63}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = \frac{- 40 b}{21} + 8 (\frac{4}{63}) + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.4

Combine $8$ and $\frac{4}{63}$ .

$2 = \frac{- 40 b}{21} + \frac{8 \cdot 4}{63} + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.5

Multiply $8$ by $4$ .

$2 = \frac{- 40 b}{21} + \frac{32}{63} + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.1.6

Move the negative in front of the fraction.

$2 = - \frac{40 b}{21} + \frac{32}{63} + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = - \frac{40 b}{21} + \frac{32}{63} + 4 b + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.2

To write $4 b$ as a fraction with a common denominator, multiply by $\frac{21}{21}$ .

$2 = - \frac{40 b}{21} + 4 b \cdot \frac{21}{21} + \frac{32}{63} + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.3

Combine $4 b$ and $\frac{21}{21}$ .

$2 = - \frac{40 b}{21} + \frac{4 b \cdot 21}{21} + \frac{32}{63} + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.4

Combine the numerators over the common denominator.

$2 = \frac{- 40 b + 4 b \cdot 21}{21} + \frac{32}{63} + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5

Find the common denominator.

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Step 2.3.4.4.1.5.1

Multiply $\frac{- 40 b + 4 b \cdot 21}{21}$ by $\frac{3}{3}$ .

$2 = \frac{- 40 b + 4 b \cdot 21}{21} \cdot \frac{3}{3} + \frac{32}{63} + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5.2

Multiply $\frac{- 40 b + 4 b \cdot 21}{21}$ by $\frac{3}{3}$ .

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{21 \cdot 3} + \frac{32}{63} + 1$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5.3

Write $1$ as a fraction with denominator $1$ .

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{21 \cdot 3} + \frac{32}{63} + \frac{1}{1}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5.4

Multiply $\frac{1}{1}$ by $\frac{63}{63}$ .

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{21 \cdot 3} + \frac{32}{63} + \frac{1}{1} \cdot \frac{63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5.5

Multiply $\frac{1}{1}$ by $\frac{63}{63}$ .

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{21 \cdot 3} + \frac{32}{63} + \frac{63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5.6

Reorder the factors of $21 \cdot 3$ .

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{3 \cdot 21} + \frac{32}{63} + \frac{63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.5.7

Multiply $3$ by $21$ .

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{63} + \frac{32}{63} + \frac{63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3}{63} + \frac{32}{63} + \frac{63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.6

Combine the numerators over the common denominator.

$2 = \frac{(- 40 b + 4 b \cdot 21) \cdot 3 + 32 + 63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.7

Simplify each term.

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Step 2.3.4.4.1.7.1

Multiply $21$ by $4$ .

$2 = \frac{(- 40 b + 84 b) \cdot 3 + 32 + 63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.7.2

Add $- 40 b$ and $84 b$ .

$2 = \frac{44 b \cdot 3 + 32 + 63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.7.3

Multiply $3$ by $44$ .

$2 = \frac{132 b + 32 + 63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = \frac{132 b + 32 + 63}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.4.1.8

Add $32$ and $63$ .

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 9 a - 3 b$

Step 2.3.4.5

Replace all occurrences of $a$ in $c = 1 - 9 a - 3 b$ with $- \frac{b}{21} + \frac{4}{315}$ .

$c = 1 - 9 (- \frac{b}{21} + \frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6

Simplify the right side.

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Step 2.3.4.6.1

Simplify $1 - 9 (- \frac{b}{21} + \frac{4}{315}) - 3 b$ .

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Step 2.3.4.6.1.1

Simplify each term.

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Step 2.3.4.6.1.1.1

Apply the distributive property.

$c = 1 - 9 (- \frac{b}{21}) - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.2

Cancel the common factor of $3$ .

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Step 2.3.4.6.1.1.2.1

Move the leading negative in $- \frac{b}{21}$ into the numerator.

$c = 1 - 9 \frac{- b}{21} - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.2.2

Factor $3$ out of $- 9$ .

$c = 1 + 3 (- 3) (\frac{- b}{21}) - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.2.3

Factor $3$ out of $21$ .

$c = 1 + 3 \cdot (- 3 \frac{- b}{3 \cdot 7}) - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.2.4

Cancel the common factor.

$c = 1 + 3 \cdot (- 3 \frac{- b}{3 \cdot 7}) - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.2.5

Rewrite the expression.

$c = 1 - 3 \frac{- b}{7} - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 - 3 \frac{- b}{7} - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.3

Combine $- 3$ and $\frac{- b}{7}$ .

$c = 1 + \frac{- 3 (- b)}{7} - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.4

Multiply $- 1$ by $- 3$ .

$c = 1 + \frac{3 b}{7} - 9 (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.5

Cancel the common factor of $9$ .

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Step 2.3.4.6.1.1.5.1

Factor $9$ out of $- 9$ .

$c = 1 + \frac{3 b}{7} + 9 (- 1) (\frac{4}{315}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.5.2

Factor $9$ out of $315$ .

$c = 1 + \frac{3 b}{7} + 9 \cdot (- 1 \frac{4}{9 \cdot 35}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.5.3

Cancel the common factor.

$c = 1 + \frac{3 b}{7} + 9 \cdot (- 1 \frac{4}{9 \cdot 35}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.5.4

Rewrite the expression.

$c = 1 + \frac{3 b}{7} - 1 (\frac{4}{35}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 + \frac{3 b}{7} - 1 (\frac{4}{35}) - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.1.6

Rewrite $- 1 (\frac{4}{35})$ as $- (\frac{4}{35})$ .

$c = 1 + \frac{3 b}{7} - \frac{4}{35} - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = 1 + \frac{3 b}{7} - \frac{4}{35} - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.2

Simplify the expression.

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Step 2.3.4.6.1.2.1

Write $1$ as a fraction with a common denominator.

$c = \frac{3 b}{7} + \frac{35}{35} - \frac{4}{35} - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.2.2

Combine the numerators over the common denominator.

$c = \frac{3 b}{7} + \frac{35 - 4}{35} - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.2.3

Subtract $4$ from $35$ .

$c = \frac{3 b}{7} + \frac{31}{35} - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{3 b}{7} + \frac{31}{35} - 3 b$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.3

To write $- 3 b$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$c = \frac{3 b}{7} - 3 b \cdot \frac{7}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.4

Simplify terms.

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Step 2.3.4.6.1.4.1

Combine $- 3 b$ and $\frac{7}{7}$ .

$c = \frac{3 b}{7} + \frac{- 3 b \cdot 7}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.4.2

Combine the numerators over the common denominator.

$c = \frac{3 b - 3 b \cdot 7}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{3 b - 3 b \cdot 7}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5

Simplify each term.

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Step 2.3.4.6.1.5.1

Simplify the numerator.

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Step 2.3.4.6.1.5.1.1

Factor $3 b$ out of $3 b - 3 b \cdot 7$ .

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Step 2.3.4.6.1.5.1.1.1

Factor $3 b$ out of $3 b$ .

$c = \frac{3 b (1) - 3 b \cdot 7}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5.1.1.2

Factor $3 b$ out of $- 3 b \cdot 7$ .

$c = \frac{3 b (1) + 3 b (- 1 \cdot 7)}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5.1.1.3

Factor $3 b$ out of $3 b (1) + 3 b (- 1 \cdot 7)$ .

$c = \frac{3 b (1 - 1 \cdot 7)}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{3 b (1 - 1 \cdot 7)}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5.1.2

Multiply $- 1$ by $7$ .

$c = \frac{3 b (1 - 7)}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5.1.3

Subtract $7$ from $1$ .

$c = \frac{3 b \cdot - 6}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5.1.4

Multiply $- 6$ by $3$ .

$c = \frac{- 18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{- 18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.4.6.1.5.2

Move the negative in front of the fraction.

$c = - \frac{18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$2 = \frac{132 b + 95}{63}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5

Solve for $b$ in $2 = \frac{132 b + 95}{63}$ .

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Step 2.3.5.1

Rewrite the equation as $\frac{132 b + 95}{63} = 2$ .

$\frac{132 b + 95}{63} = 2$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.2

Multiply both sides by $63$ .

$\frac{132 b + 95}{63} \cdot 63 = 2 \cdot 63$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.3

Simplify.

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Step 2.3.5.3.1

Simplify the left side.

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Step 2.3.5.3.1.1

Cancel the common factor of $63$ .

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Step 2.3.5.3.1.1.1

Cancel the common factor.

$\frac{132 b + 95}{63} \cdot 63 = 2 \cdot 63$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.3.1.1.2

Rewrite the expression.

$132 b + 95 = 2 \cdot 63$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$132 b + 95 = 2 \cdot 63$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$132 b + 95 = 2 \cdot 63$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.3.2

Simplify the right side.

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Step 2.3.5.3.2.1

Multiply $2$ by $63$ .

$132 b + 95 = 126$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$132 b + 95 = 126$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$132 b + 95 = 126$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.4

Solve for $b$ .

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Step 2.3.5.4.1

Move all terms not containing $b$ to the right side of the equation.

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Step 2.3.5.4.1.1

Subtract $95$ from both sides of the equation.

$132 b = 126 - 95$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.4.1.2

Subtract $95$ from $126$ .

$132 b = 31$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$132 b = 31$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.4.2

Divide each term in $132 b = 31$ by $132$ and simplify.

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Step 2.3.5.4.2.1

Divide each term in $132 b = 31$ by $132$ .

$\frac{132 b}{132} = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.4.2.2

Simplify the left side.

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Step 2.3.5.4.2.2.1

Cancel the common factor of $132$ .

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Step 2.3.5.4.2.2.1.1

Cancel the common factor.

$\frac{132 b}{132} = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.5.4.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$b = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$b = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$b = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$b = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$b = \frac{31}{132}$

$c = - \frac{18 b}{7} + \frac{31}{35}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6

Replace all occurrences of $b$ with $\frac{31}{132}$ in each equation.

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Step 2.3.6.1

Replace all occurrences of $b$ in $c = - \frac{18 b}{7} + \frac{31}{35}$ with $\frac{31}{132}$ .

$c = - \frac{18 (\frac{31}{132})}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2

Simplify the right side.

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Step 2.3.6.2.1

Simplify $- \frac{18 (\frac{31}{132})}{7} + \frac{31}{35}$ .

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Step 2.3.6.2.1.1

Simplify each term.

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Step 2.3.6.2.1.1.1

Combine $18$ and $\frac{31}{132}$ .

$c = - \frac{\frac{18 \cdot 31}{132}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.2

Multiply $18$ by $31$ .

$c = - \frac{\frac{558}{132}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.3

Cancel the common factor of $558$ and $132$ .

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Step 2.3.6.2.1.1.3.1

Factor $6$ out of $558$ .

$c = - \frac{\frac{6 (93)}{132}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.3.2

Cancel the common factors.

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Step 2.3.6.2.1.1.3.2.1

Factor $6$ out of $132$ .

$c = - \frac{\frac{6 \cdot 93}{6 \cdot 22}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.3.2.2

Cancel the common factor.

$c = - \frac{\frac{6 \cdot 93}{6 \cdot 22}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.3.2.3

Rewrite the expression.

$c = - \frac{\frac{93}{22}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{\frac{93}{22}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{\frac{93}{22}}{7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.4

Multiply the numerator by the reciprocal of the denominator.

$c = - (\frac{93}{22} \cdot \frac{1}{7}) + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.5

Multiply $\frac{93}{22} \cdot \frac{1}{7}$ .

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Step 2.3.6.2.1.1.5.1

Multiply $\frac{93}{22}$ by $\frac{1}{7}$ .

$c = - \frac{93}{22 \cdot 7} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.1.5.2

Multiply $22$ by $7$ .

$c = - \frac{93}{154} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{93}{154} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{93}{154} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.2

To write $- \frac{93}{154}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$c = - \frac{93}{154} \cdot \frac{5}{5} + \frac{31}{35}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.3

To write $\frac{31}{35}$ as a fraction with a common denominator, multiply by $\frac{22}{22}$ .

$c = - \frac{93}{154} \cdot \frac{5}{5} + \frac{31}{35} \cdot \frac{22}{22}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.4

Write each expression with a common denominator of $770$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.6.2.1.4.1

Multiply $\frac{93}{154}$ by $\frac{5}{5}$ .

$c = - \frac{93 \cdot 5}{154 \cdot 5} + \frac{31}{35} \cdot \frac{22}{22}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.4.2

Multiply $154$ by $5$ .

$c = - \frac{93 \cdot 5}{770} + \frac{31}{35} \cdot \frac{22}{22}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.4.3

Multiply $\frac{31}{35}$ by $\frac{22}{22}$ .

$c = - \frac{93 \cdot 5}{770} + \frac{31 \cdot 22}{35 \cdot 22}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.4.4

Multiply $35$ by $22$ .

$c = - \frac{93 \cdot 5}{770} + \frac{31 \cdot 22}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = - \frac{93 \cdot 5}{770} + \frac{31 \cdot 22}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.5

Combine the numerators over the common denominator.

$c = \frac{- 93 \cdot 5 + 31 \cdot 22}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.6

Simplify the numerator.

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Step 2.3.6.2.1.6.1

Multiply $- 93$ by $5$ .

$c = \frac{- 465 + 31 \cdot 22}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.6.2

Multiply $31$ by $22$ .

$c = \frac{- 465 + 682}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.6.3

Add $- 465$ and $682$ .

$c = \frac{217}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{217}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.7

Cancel the common factor of $217$ and $770$ .

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Step 2.3.6.2.1.7.1

Factor $7$ out of $217$ .

$c = \frac{7 (31)}{770}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.7.2

Cancel the common factors.

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Step 2.3.6.2.1.7.2.1

Factor $7$ out of $770$ .

$c = \frac{7 \cdot 31}{7 \cdot 110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.7.2.2

Cancel the common factor.

$c = \frac{7 \cdot 31}{7 \cdot 110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.2.1.7.2.3

Rewrite the expression.

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$3 = \frac{120 b + 109}{45}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.3

Replace all occurrences of $b$ in $3 = \frac{120 b + 109}{45}$ with $\frac{31}{132}$ .

$3 = \frac{120 (\frac{31}{132}) + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4

Simplify the right side.

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Step 2.3.6.4.1

Simplify $\frac{120 (\frac{31}{132}) + 109}{45}$ .

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Step 2.3.6.4.1.1

Simplify the numerator.

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Step 2.3.6.4.1.1.1

Cancel the common factor of $12$ .

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Step 2.3.6.4.1.1.1.1

Factor $12$ out of $120$ .

$3 = \frac{12 (10) (\frac{31}{132}) + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.1.2

Factor $12$ out of $132$ .

$3 = \frac{12 \cdot (10 (\frac{31}{12 \cdot 11})) + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.1.3

Cancel the common factor.

$3 = \frac{12 \cdot (10 (\frac{31}{12 \cdot 11})) + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.1.4

Rewrite the expression.

$3 = \frac{10 (\frac{31}{11}) + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = \frac{10 (\frac{31}{11}) + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.2

Combine $10$ and $\frac{31}{11}$ .

$3 = \frac{\frac{10 \cdot 31}{11} + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.3

Multiply $10$ by $31$ .

$3 = \frac{\frac{310}{11} + 109}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.4

To write $109$ as a fraction with a common denominator, multiply by $\frac{11}{11}$ .

$3 = \frac{\frac{310}{11} + 109 \cdot \frac{11}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.5

Combine $109$ and $\frac{11}{11}$ .

$3 = \frac{\frac{310}{11} + \frac{109 \cdot 11}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.6

Combine the numerators over the common denominator.

$3 = \frac{\frac{310 + 109 \cdot 11}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.7

Simplify the numerator.

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Step 2.3.6.4.1.1.7.1

Multiply $109$ by $11$ .

$3 = \frac{\frac{310 + 1199}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.1.7.2

Add $310$ and $1199$ .

$3 = \frac{\frac{1509}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = \frac{\frac{1509}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = \frac{\frac{1509}{11}}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.2

Multiply the numerator by the reciprocal of the denominator.

$3 = \frac{1509}{11} \cdot \frac{1}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.3

Cancel the common factor of $3$ .

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Step 2.3.6.4.1.3.1

Factor $3$ out of $1509$ .

$3 = \frac{3 (503)}{11} \cdot \frac{1}{45}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.3.2

Factor $3$ out of $45$ .

$3 = \frac{3 \cdot 503}{11} \cdot \frac{1}{3 \cdot 15}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.3.3

Cancel the common factor.

$3 = \frac{3 \cdot 503}{11} \cdot \frac{1}{3 \cdot 15}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.3.4

Rewrite the expression.

$3 = \frac{503}{11} \cdot \frac{1}{15}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = \frac{503}{11} \cdot \frac{1}{15}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.4

Multiply $\frac{503}{11}$ by $\frac{1}{15}$ .

$3 = \frac{503}{11 \cdot 15}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.4.1.5

Multiply $11$ by $15$ .

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{b}{21} + \frac{4}{315}$

Step 2.3.6.5

Replace all occurrences of $b$ in $a = - \frac{b}{21} + \frac{4}{315}$ with $\frac{31}{132}$ .

$a = - \frac{\frac{31}{132}}{21} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6

Simplify the right side.

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Step 2.3.6.6.1

Simplify $- \frac{\frac{31}{132}}{21} + \frac{4}{315}$ .

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Step 2.3.6.6.1.1

Simplify each term.

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Step 2.3.6.6.1.1.1

Multiply the numerator by the reciprocal of the denominator.

$a = - (\frac{31}{132} \cdot \frac{1}{21}) + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.1.2

Multiply $\frac{31}{132} \cdot \frac{1}{21}$ .

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Step 2.3.6.6.1.1.2.1

Multiply $\frac{31}{132}$ by $\frac{1}{21}$ .

$a = - \frac{31}{132 \cdot 21} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.1.2.2

Multiply $132$ by $21$ .

$a = - \frac{31}{2772} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{31}{2772} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{31}{2772} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.2

To write $- \frac{31}{2772}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$a = - \frac{31}{2772} \cdot \frac{5}{5} + \frac{4}{315}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.3

To write $\frac{4}{315}$ as a fraction with a common denominator, multiply by $\frac{44}{44}$ .

$a = - \frac{31}{2772} \cdot \frac{5}{5} + \frac{4}{315} \cdot \frac{44}{44}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.4

Write each expression with a common denominator of $13860$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.6.6.1.4.1

Multiply $\frac{31}{2772}$ by $\frac{5}{5}$ .

$a = - \frac{31 \cdot 5}{2772 \cdot 5} + \frac{4}{315} \cdot \frac{44}{44}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.4.2

Multiply $2772$ by $5$ .

$a = - \frac{31 \cdot 5}{13860} + \frac{4}{315} \cdot \frac{44}{44}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.4.3

Multiply $\frac{4}{315}$ by $\frac{44}{44}$ .

$a = - \frac{31 \cdot 5}{13860} + \frac{4 \cdot 44}{315 \cdot 44}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.4.4

Multiply $315$ by $44$ .

$a = - \frac{31 \cdot 5}{13860} + \frac{4 \cdot 44}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = - \frac{31 \cdot 5}{13860} + \frac{4 \cdot 44}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.5

Combine the numerators over the common denominator.

$a = \frac{- 31 \cdot 5 + 4 \cdot 44}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.6

Simplify the numerator.

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Step 2.3.6.6.1.6.1

Multiply $- 31$ by $5$ .

$a = \frac{- 155 + 4 \cdot 44}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.6.2

Multiply $4$ by $44$ .

$a = \frac{- 155 + 176}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.6.3

Add $- 155$ and $176$ .

$a = \frac{21}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = \frac{21}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.7

Cancel the common factor of $21$ and $13860$ .

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Step 2.3.6.6.1.7.1

Factor $21$ out of $21$ .

$a = \frac{21 (1)}{13860}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.7.2

Cancel the common factors.

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Step 2.3.6.6.1.7.2.1

Factor $21$ out of $13860$ .

$a = \frac{21 \cdot 1}{21 \cdot 660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.7.2.2

Cancel the common factor.

$a = \frac{21 \cdot 1}{21 \cdot 660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.6.6.1.7.2.3

Rewrite the expression.

$a = \frac{1}{660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = \frac{1}{660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = \frac{1}{660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = \frac{1}{660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = \frac{1}{660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

$a = \frac{1}{660}$

$3 = \frac{503}{165}$

$c = \frac{31}{110}$

$b = \frac{31}{132}$

Step 2.3.7

Since $3 = \frac{503}{165}$ is not true, there is no solution.

No solution

Step 2.4

Calculate the value of $y$ using each $x$ value in the table and compare this value to the given $y$ value in the table.

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Step 2.4.1

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = 0$ , $b = 0$ , $c = 0$ , and $x = 3$ .

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Step 2.4.1.1

Simplify each term.

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Step 2.4.1.1.1

Raise $3$ to the power of $2$ .

$y = 0 \cdot 9 + (0) \cdot (3) + 0$

Step 2.4.1.1.2

Multiply $0$ by $9$ .

$y = 0 + (0) \cdot (3) + 0$

Step 2.4.1.1.3

Multiply $0$ by $3$ .

$y = 0 + 0 + 0$

Step 2.4.1.2

Simplify by adding numbers.

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Step 2.4.1.2.1

Add $0$ and $0$ .

$y = 0 + 0$

Step 2.4.1.2.2

Add $0$ and $0$ .

$y = 0$

Step 2.4.2

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check does not pass, since $y = 0$ and $y = 1$ . The function rule can't be quadratic.

$0 \neq 1$

Step 2.4.3

Since $y \neq y$ for the corresponding $x$ values, the function is not quadratic.

The function is not quadratic

Step 3

Check if the function rule is cubic.

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Step 3.1

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{3} + b x^{2} + c x + d$ .

$y = a x^{3} + b x^{2} + c x + d$

Step 3.2

Build a set of $4$ equations from the table such that $y = a x^{3} + b x^{2} + c x + d$ .

Step 3.3

Calculate the values of $a$ , $b$ , $c$ , and $d$ .

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Step 3.3.1

Solve for $d$ in $1 = a \cdot 3^{3} + b \cdot 3^{2} + c (3) + d$ .

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Step 3.3.1.1

Rewrite the equation as $a \cdot 3^{3} + b \cdot 3^{2} + c (3) + d = 1$ .

$a \cdot 3^{3} + b \cdot 3^{2} + c (3) + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.2

Simplify each term.

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Step 3.3.1.2.1

Raise $3$ to the power of $3$ .

$a \cdot 27 + b \cdot 3^{2} + c (3) + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.2.2

Move $27$ to the left of $a$ .

$27 \cdot a + b \cdot 3^{2} + c (3) + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.2.3

Raise $3$ to the power of $2$ .

$27 a + b \cdot 9 + c (3) + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.2.4

Move $9$ to the left of $b$ .

$27 a + 9 \cdot b + c (3) + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.2.5

Move $3$ to the left of $c$ .

$27 a + 9 b + 3 c + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$27 a + 9 b + 3 c + d = 1$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.3

Move all terms not containing $d$ to the right side of the equation.

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Step 3.3.1.3.1

Subtract $27 a$ from both sides of the equation.

$9 b + 3 c + d = 1 - 27 a$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.3.2

Subtract $9 b$ from both sides of the equation.

$3 c + d = 1 - 27 a - 9 b$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.1.3.3

Subtract $3 c$ from both sides of the equation.

$d = 1 - 27 a - 9 b - 3 c$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$d = 1 - 27 a - 9 b - 3 c$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$d = 1 - 27 a - 9 b - 3 c$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2

Replace all occurrences of $d$ with $1 - 27 a - 9 b - 3 c$ in each equation.

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Step 3.3.2.1

Replace all occurrences of $d$ in $2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + d$ with $1 - 27 a - 9 b - 3 c$ .

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2

Simplify $2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.2.1

Simplify the left side.

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Step 3.3.2.2.1.1

Remove parentheses.

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2

Simplify the right side.

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Step 3.3.2.2.2.1

Simplify $a \cdot 7^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.2.2.1.1

Combine the opposite terms in $a \cdot 7^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.2.2.1.1.1

Reorder the factors in the terms $c (3)$ and $- 3 c$ .

$2 = a \cdot 7^{3} + b \cdot 3^{2} + 3 c + 1 - 27 a - 9 b - 3 c$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.1.2

Subtract $3 c$ from $3 c$ .

$2 = a \cdot 7^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b + 0$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.1.3

Add $a \cdot 7^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$ and $0$ .

$2 = a \cdot 7^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = a \cdot 7^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.2

Simplify each term.

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Step 3.3.2.2.2.1.2.1

Raise $7$ to the power of $3$ .

$2 = a \cdot 343 + b \cdot 3^{2} + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.2.2

Move $343$ to the left of $a$ .

$2 = 343 \cdot a + b \cdot 3^{2} + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.2.3

Raise $3$ to the power of $2$ .

$2 = 343 a + b \cdot 9 + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.2.4

Move $9$ to the left of $b$ .

$2 = 343 a + 9 b + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = 343 a + 9 b + 1 - 27 a - 9 b$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.3

Simplify by adding terms.

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Step 3.3.2.2.2.1.3.1

Combine the opposite terms in $343 a + 9 b + 1 - 27 a - 9 b$ .

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Step 3.3.2.2.2.1.3.1.1

Subtract $9 b$ from $9 b$ .

$2 = 343 a + 1 - 27 a + 0$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.3.1.2

Add $343 a + 1 - 27 a$ and $0$ .

$2 = 343 a + 1 - 27 a$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = 343 a + 1 - 27 a$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.2.2.1.3.2

Subtract $27 a$ from $343 a$ .

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.3

Replace all occurrences of $d$ in $3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + d$ with $1 - 27 a - 9 b - 3 c$ .

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4

Simplify $3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.4.1

Simplify the left side.

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Step 3.3.2.4.1.1

Remove parentheses.

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2

Simplify the right side.

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Step 3.3.2.4.2.1

Simplify $a \cdot 11^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.4.2.1.1

Combine the opposite terms in $a \cdot 11^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.4.2.1.1.1

Reorder the factors in the terms $c (3)$ and $- 3 c$ .

$3 = a \cdot 11^{3} + b \cdot 3^{2} + 3 c + 1 - 27 a - 9 b - 3 c$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.1.2

Subtract $3 c$ from $3 c$ .

$3 = a \cdot 11^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b + 0$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.1.3

Add $a \cdot 11^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$ and $0$ .

$3 = a \cdot 11^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = a \cdot 11^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.2

Simplify each term.

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Step 3.3.2.4.2.1.2.1

Raise $11$ to the power of $3$ .

$3 = a \cdot 1331 + b \cdot 3^{2} + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.2.2

Move $1331$ to the left of $a$ .

$3 = 1331 \cdot a + b \cdot 3^{2} + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.2.3

Raise $3$ to the power of $2$ .

$3 = 1331 a + b \cdot 9 + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.2.4

Move $9$ to the left of $b$ .

$3 = 1331 a + 9 b + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = 1331 a + 9 b + 1 - 27 a - 9 b$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.3

Simplify by adding terms.

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Step 3.3.2.4.2.1.3.1

Combine the opposite terms in $1331 a + 9 b + 1 - 27 a - 9 b$ .

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Step 3.3.2.4.2.1.3.1.1

Subtract $9 b$ from $9 b$ .

$3 = 1331 a + 1 - 27 a + 0$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.3.1.2

Add $1331 a + 1 - 27 a$ and $0$ .

$3 = 1331 a + 1 - 27 a$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = 1331 a + 1 - 27 a$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.4.2.1.3.2

Subtract $27 a$ from $1331 a$ .

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$

Step 3.3.2.5

Replace all occurrences of $d$ in $5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + d$ with $1 - 27 a - 9 b - 3 c$ .

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6

Simplify $5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.6.1

Simplify the left side.

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Step 3.3.2.6.1.1

Remove parentheses.

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2

Simplify the right side.

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Step 3.3.2.6.2.1

Simplify $a \cdot 18^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.6.2.1.1

Combine the opposite terms in $a \cdot 18^{3} + b \cdot 3^{2} + c (3) + 1 - 27 a - 9 b - 3 c$ .

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Step 3.3.2.6.2.1.1.1

Reorder the factors in the terms $c (3)$ and $- 3 c$ .

$5 = a \cdot 18^{3} + b \cdot 3^{2} + 3 c + 1 - 27 a - 9 b - 3 c$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.1.2

Subtract $3 c$ from $3 c$ .

$5 = a \cdot 18^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b + 0$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.1.3

Add $a \cdot 18^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$ and $0$ .

$5 = a \cdot 18^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = a \cdot 18^{3} + b \cdot 3^{2} + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.2

Simplify each term.

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Step 3.3.2.6.2.1.2.1

Raise $18$ to the power of $3$ .

$5 = a \cdot 5832 + b \cdot 3^{2} + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.2.2

Move $5832$ to the left of $a$ .

$5 = 5832 \cdot a + b \cdot 3^{2} + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.2.3

Raise $3$ to the power of $2$ .

$5 = 5832 a + b \cdot 9 + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.2.4

Move $9$ to the left of $b$ .

$5 = 5832 a + 9 b + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5832 a + 9 b + 1 - 27 a - 9 b$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.3

Simplify by adding terms.

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Step 3.3.2.6.2.1.3.1

Combine the opposite terms in $5832 a + 9 b + 1 - 27 a - 9 b$ .

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Step 3.3.2.6.2.1.3.1.1

Subtract $9 b$ from $9 b$ .

$5 = 5832 a + 1 - 27 a + 0$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.3.1.2

Add $5832 a + 1 - 27 a$ and $0$ .

$5 = 5832 a + 1 - 27 a$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5832 a + 1 - 27 a$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.2.6.2.1.3.2

Subtract $27 a$ from $5832 a$ .

$5 = 5805 a + 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5805 a + 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5805 a + 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5805 a + 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5805 a + 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5 = 5805 a + 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.3

Solve for $a$ in $5 = 5805 a + 1$ .

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Step 3.3.3.1

Rewrite the equation as $5805 a + 1 = 5$ .

$5805 a + 1 = 5$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.3.2

Move all terms not containing $a$ to the right side of the equation.

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Step 3.3.3.2.1

Subtract $1$ from both sides of the equation.

$5805 a = 5 - 1$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.3.2.2

Subtract $1$ from $5$ .

$5805 a = 4$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$5805 a = 4$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.3.3

Divide each term in $5805 a = 4$ by $5805$ and simplify.

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Step 3.3.3.3.1

Divide each term in $5805 a = 4$ by $5805$ .

$\frac{5805 a}{5805} = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.3.3.2

Simplify the left side.

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Step 3.3.3.3.2.1

Cancel the common factor of $5805$ .

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Step 3.3.3.3.2.1.1

Cancel the common factor.

$\frac{5805 a}{5805} = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$a = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$a = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$a = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$a = \frac{4}{5805}$

$3 = 1304 a + 1$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4

Replace all occurrences of $a$ with $\frac{4}{5805}$ in each equation.

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Step 3.3.4.1

Replace all occurrences of $a$ in $3 = 1304 a + 1$ with $\frac{4}{5805}$ .

$3 = 1304 (\frac{4}{5805}) + 1$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.2

Simplify the right side.

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Step 3.3.4.2.1

Simplify $1304 (\frac{4}{5805}) + 1$ .

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Step 3.3.4.2.1.1

Multiply $1304 (\frac{4}{5805})$ .

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Step 3.3.4.2.1.1.1

Combine $1304$ and $\frac{4}{5805}$ .

$3 = \frac{1304 \cdot 4}{5805} + 1$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.2.1.1.2

Multiply $1304$ by $4$ .

$3 = \frac{5216}{5805} + 1$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = \frac{5216}{5805} + 1$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.2.1.2

Write $1$ as a fraction with a common denominator.

$3 = \frac{5216}{5805} + \frac{5805}{5805}$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.2.1.3

Combine the numerators over the common denominator.

$3 = \frac{5216 + 5805}{5805}$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.2.1.4

Add $5216$ and $5805$ .

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$2 = 316 a + 1$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.3

Replace all occurrences of $a$ in $2 = 316 a + 1$ with $\frac{4}{5805}$ .

$2 = 316 (\frac{4}{5805}) + 1$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.4

Simplify the right side.

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Step 3.3.4.4.1

Simplify $316 (\frac{4}{5805}) + 1$ .

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Step 3.3.4.4.1.1

Multiply $316 (\frac{4}{5805})$ .

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Step 3.3.4.4.1.1.1

Combine $316$ and $\frac{4}{5805}$ .

$2 = \frac{316 \cdot 4}{5805} + 1$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.4.1.1.2

Multiply $316$ by $4$ .

$2 = \frac{1264}{5805} + 1$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

$2 = \frac{1264}{5805} + 1$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.4.1.2

Write $1$ as a fraction with a common denominator.

$2 = \frac{1264}{5805} + \frac{5805}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.4.1.3

Combine the numerators over the common denominator.

$2 = \frac{1264 + 5805}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.4.1.4

Add $1264$ and $5805$ .

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 27 a - 9 b - 3 c$

Step 3.3.4.5

Replace all occurrences of $a$ in $d = 1 - 27 a - 9 b - 3 c$ with $\frac{4}{5805}$ .

$d = 1 - 27 (\frac{4}{5805}) - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6

Simplify the right side.

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Step 3.3.4.6.1

Simplify $1 - 27 (\frac{4}{5805}) - 9 b - 3 c$ .

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Step 3.3.4.6.1.1

Simplify each term.

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Step 3.3.4.6.1.1.1

Cancel the common factor of $27$ .

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Step 3.3.4.6.1.1.1.1

Factor $27$ out of $- 27$ .

$d = 1 + 27 (- 1) (\frac{4}{5805}) - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.1.1.2

Factor $27$ out of $5805$ .

$d = 1 + 27 \cdot (- 1 \frac{4}{27 \cdot 215}) - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.1.1.3

Cancel the common factor.

$d = 1 + 27 \cdot (- 1 \frac{4}{27 \cdot 215}) - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.1.1.4

Rewrite the expression.

$d = 1 - 1 (\frac{4}{215}) - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - 1 (\frac{4}{215}) - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.1.2

Rewrite $- 1 (\frac{4}{215})$ as $- (\frac{4}{215})$ .

$d = 1 - \frac{4}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = 1 - \frac{4}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.2

Simplify the expression.

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Step 3.3.4.6.1.2.1

Write $1$ as a fraction with a common denominator.

$d = \frac{215}{215} - \frac{4}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.2.2

Combine the numerators over the common denominator.

$d = \frac{215 - 4}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.4.6.1.2.3

Subtract $4$ from $215$ .

$d = \frac{211}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = \frac{211}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = \frac{211}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = \frac{211}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

$d = \frac{211}{215} - 9 b - 3 c$

$2 = \frac{7069}{5805}$

$3 = \frac{11021}{5805}$

$a = \frac{4}{5805}$

Step 3.3.5

Since $2 = \frac{7069}{5805}$ is not true, there is no solution.

No solution

Step 3.4

Calculate the value of $y$ using each $x$ value in the table and compare this value to the given $y$ value in the table.

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Step 3.4.1

Calculate the value of $y$ such that $y = a x^{3} + b$ when $a = 0$ , $b = 0$ , $c = 0$ , $d = 0$ , and $x = 3$ .

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Step 3.4.1.1

Simplify each term.

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Step 3.4.1.1.1

Raise $3$ to the power of $3$ .

$y = 0 \cdot 27 + (0) \cdot (3^{2}) + (0) \cdot ((3) \cdot 0)$

Step 3.4.1.1.2

Multiply $0$ by $27$ .

$y = 0 + (0) \cdot (3^{2}) + (0) \cdot ((3) \cdot 0)$

Step 3.4.1.1.3

Raise $3$ to the power of $2$ .

$y = 0 + 0 \cdot 9 + (0) \cdot ((3) \cdot 0)$

Step 3.4.1.1.4

Multiply $0$ by $9$ .

$y = 0 + 0 + (0) \cdot ((3) \cdot 0)$

Step 3.4.1.1.5

Multiply $(0) ((3) \cdot 0)$ .

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Step 3.4.1.1.5.1

Multiply $3$ by $0$ .

$y = 0 + 0 + 0 \cdot 0$

Step 3.4.1.1.5.2

Multiply $0$ by $0$ .

$y = 0 + 0 + 0$

Step 3.4.1.2

Simplify by adding numbers.

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Step 3.4.1.2.1

Add $0$ and $0$ .

$y = 0 + 0$

Step 3.4.1.2.2

Add $0$ and $0$ .

$y = 0$

Step 3.4.2

If the table has a cubic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check does not pass, since $y = 0$ and $y = 1$ . The function rule can't be cubic.

$0 \neq 1$

Step 3.4.3

Since $y \neq y$ for the corresponding $x$ values, the function is not cubic.

The function is not cubic

Step 4

There are no values for $a$ , $b$ , $c$ , and $d$ in the equations $y = a x + b$ , $y = a x^{2} + b x + c$ , and $y = a x^{3} + b x^{2} + c x + d$ that work for every pair of $x$ and $y$ .

The table does not have a function rule that is linear, quadratic, or cubic.