Algebra Examples

$(- 6, - 1)$ , $(- 3, 2)$ , $(- 1, 4)$ , $(2, 7)$

Step 1

To find if the relation is linear, check to see if the values follow the linear form $y = a x + b$ .

$y = a x + b$

Step 2

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

$- 1 = a (- 6) + b 2 = a (- 3) + b 4 = a (- 1) + b 7 = a (2) + b$

Step 3

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

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Step 3.1

Solve for $b$ in $- 1 = a (- 6) + b$ .

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Step 3.1.1

Rewrite the equation as $a (- 6) + b = - 1$ .

$a (- 6) + b = - 1$

$2 = a (- 3) + b$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.1.2

Move $- 6$ to the left of $a$ .

$- 6 a + b = - 1$

$2 = a (- 3) + b$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.1.3

Add $6 a$ to both sides of the equation.

$b = - 1 + 6 a$

$2 = a (- 3) + b$

$4 = a (- 1) + b$

$7 = a (2) + b$

$b = - 1 + 6 a$

$2 = a (- 3) + b$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.2

Replace all occurrences of $b$ with $- 1 + 6 a$ in each equation.

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Step 3.2.1

Replace all occurrences of $b$ in $2 = a (- 3) + b$ with $- 1 + 6 a$ .

$2 = a (- 3) - 1 + 6 a$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.2.2

Simplify $2 = a (- 3) - 1 + 6 a$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

$2 = a (- 3) - 1 + 6 a$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

$2 = a (- 3) - 1 + 6 a$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify $a (- 3) - 1 + 6 a$ .

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Step 3.2.2.2.1.1

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

$2 = - 3 a - 1 + 6 a$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.2.2.2.1.2

Add $- 3 a$ and $6 a$ .

$2 = 3 a - 1$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$4 = a (- 1) + b$

$7 = a (2) + b$

Step 3.2.3

Replace all occurrences of $b$ in $4 = a (- 1) + b$ with $- 1 + 6 a$ .

$4 = a (- 1) - 1 + 6 a$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

Step 3.2.4

Simplify $4 = a (- 1) - 1 + 6 a$ .

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Step 3.2.4.1

Simplify the left side.

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Step 3.2.4.1.1

Remove parentheses.

$4 = a (- 1) - 1 + 6 a$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

$4 = a (- 1) - 1 + 6 a$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

Step 3.2.4.2

Simplify the right side.

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Step 3.2.4.2.1

Simplify $a (- 1) - 1 + 6 a$ .

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Step 3.2.4.2.1.1

Simplify each term.

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Step 3.2.4.2.1.1.1

Move $- 1$ to the left of $a$ .

$4 = - 1 a - 1 + 6 a$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

Step 3.2.4.2.1.1.2

Rewrite $- 1 a$ as $- a$ .

$4 = - a - 1 + 6 a$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

$4 = - a - 1 + 6 a$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

Step 3.2.4.2.1.2

Add $- a$ and $6 a$ .

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) + b$

Step 3.2.5

Replace all occurrences of $b$ in $7 = a (2) + b$ with $- 1 + 6 a$ .

$7 = a (2) - 1 + 6 a$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.2.6

Simplify $7 = a (2) - 1 + 6 a$ .

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Step 3.2.6.1

Simplify the left side.

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Step 3.2.6.1.1

Remove parentheses.

$7 = a (2) - 1 + 6 a$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = a (2) - 1 + 6 a$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.2.6.2

Simplify the right side.

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Step 3.2.6.2.1

Simplify $a (2) - 1 + 6 a$ .

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Step 3.2.6.2.1.1

Move $2$ to the left of $a$ .

$7 = 2 a - 1 + 6 a$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.2.6.2.1.2

Add $2 a$ and $6 a$ .

$7 = 8 a - 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = 8 a - 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = 8 a - 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = 8 a - 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$7 = 8 a - 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3

Solve for $a$ in $7 = 8 a - 1$ .

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Step 3.3.1

Rewrite the equation as $8 a - 1 = 7$ .

$8 a - 1 = 7$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3.2

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

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Step 3.3.2.1

Add $1$ to both sides of the equation.

$8 a = 7 + 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3.2.2

Add $7$ and $1$ .

$8 a = 8$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$8 a = 8$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3.3

Divide each term in $8 a = 8$ by $8$ and simplify.

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Step 3.3.3.1

Divide each term in $8 a = 8$ by $8$ .

$\frac{8 a}{8} = \frac{8}{8}$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3.3.2

Simplify the left side.

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Step 3.3.3.2.1

Cancel the common factor of $8$ .

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Step 3.3.3.2.1.1

Cancel the common factor.

$\frac{8 a}{8} = \frac{8}{8}$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{8}{8}$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$a = \frac{8}{8}$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$a = \frac{8}{8}$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.3.3.3

Simplify the right side.

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Step 3.3.3.3.1

Divide $8$ by $8$ .

$a = 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$a = 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$a = 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$a = 1$

$4 = 5 a - 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.4

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

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Step 3.4.1

Replace all occurrences of $a$ in $4 = 5 a - 1$ with $1$ .

$4 = 5 (1) - 1$

$a = 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify $5 (1) - 1$ .

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Step 3.4.2.1.1

Multiply $5$ by $1$ .

$4 = 5 - 1$

$a = 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.4.2.1.2

Subtract $1$ from $5$ .

$4 = 4$

$a = 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$4 = 4$

$a = 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

$4 = 4$

$a = 1$

$2 = 3 a - 1$

$b = - 1 + 6 a$

Step 3.4.3

Replace all occurrences of $a$ in $2 = 3 a - 1$ with $1$ .

$2 = 3 (1) - 1$

$4 = 4$

$a = 1$

$b = - 1 + 6 a$

Step 3.4.4

Simplify the right side.

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Step 3.4.4.1

Simplify $3 (1) - 1$ .

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Step 3.4.4.1.1

Multiply $3$ by $1$ .

$2 = 3 - 1$

$4 = 4$

$a = 1$

$b = - 1 + 6 a$

Step 3.4.4.1.2

Subtract $1$ from $3$ .

$2 = 2$

$4 = 4$

$a = 1$

$b = - 1 + 6 a$

$2 = 2$

$4 = 4$

$a = 1$

$b = - 1 + 6 a$

$2 = 2$

$4 = 4$

$a = 1$

$b = - 1 + 6 a$

Step 3.4.5

Replace all occurrences of $a$ in $b = - 1 + 6 a$ with $1$ .

$b = - 1 + 6 (1)$

$2 = 2$

$4 = 4$

$a = 1$

Step 3.4.6

Simplify the right side.

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Step 3.4.6.1

Simplify $- 1 + 6 (1)$ .

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Step 3.4.6.1.1

Multiply $6$ by $1$ .

$b = - 1 + 6$

$2 = 2$

$4 = 4$

$a = 1$

Step 3.4.6.1.2

Add $- 1$ and $6$ .

$b = 5$

$2 = 2$

$4 = 4$

$a = 1$

$b = 5$

$2 = 2$

$4 = 4$

$a = 1$

$b = 5$

$2 = 2$

$4 = 4$

$a = 1$

$b = 5$

$2 = 2$

$4 = 4$

$a = 1$

Step 3.5

Remove any equations from the system that are always true.

$b = 5$

$a = 1$

Step 3.6

List all of the solutions.

$b = 5, a = 1$

Step 4

Calculate the value of $y$ using each $x$ value in the relation and compare this value to the given $y$ value in the relation.

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Step 4.1

Calculate the value of $y$ when $a = 1$ , $b = 5$ , and $x = - 6$ .

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Step 4.1.1

Multiply $- 6$ by $1$ .

$y = - 6 + 5$

Step 4.1.2

Add $- 6$ and $5$ .

$y = - 1$

Step 4.2

If the relation is linear, $y = y$ for the corresponding $x$ value, $x = - 6$ . This check passes since $y = - 1$ and $y = - 1$ .

$- 1 = - 1$

Step 4.3

Calculate the value of $y$ when $a = 1$ , $b = 5$ , and $x = - 3$ .

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Step 4.3.1

Multiply $- 3$ by $1$ .

$y = - 3 + 5$

Step 4.3.2

Add $- 3$ and $5$ .

$y = 2$

Step 4.4

If the relation is linear, $y = y$ for the corresponding $x$ value, $x = - 3$ . This check passes since $y = 2$ and $y = 2$ .

$2 = 2$

Step 4.5

Calculate the value of $y$ when $a = 1$ , $b = 5$ , and $x = - 1$ .

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Step 4.5.1

Multiply $- 1$ by $1$ .

$y = - 1 + 5$

Step 4.5.2

Add $- 1$ and $5$ .

$y = 4$

Step 4.6

If the relation is linear, $y = y$ for the corresponding $x$ value, $x = - 1$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

Step 4.7

Calculate the value of $y$ when $a = 1$ , $b = 5$ , and $x = 2$ .

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Step 4.7.1

Multiply $2$ by $1$ .

$y = 2 + 5$

Step 4.7.2

Add $2$ and $5$ .

$y = 7$

Step 4.8

If the relation is linear, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 7$ and $y = 7$ .

$7 = 7$

Step 4.9

Since $y = y$ for the corresponding $x$ values, the relation is linear.

The relation is linear