Precalculus Examples

$[\begin{array}{c} 1 & 2 \\ - 2 & - 3 \end{array}] = [\begin{array}{c} x & 2 \\ - 2 & - 3 \end{array}]$

Step 1

Find the function rule.

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Step 1.1

Check if the function rule is linear.

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Step 1.1.1

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 1.1.2

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

$2 = a (2) + b - 2 = a (- 2) + b - 3 = a (- 3) + b$

Step 1.1.3

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

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Step 1.1.3.1

Solve for $b$ in $2 = a (2) + b$ .

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Step 1.1.3.1.1

Rewrite the equation as $a (2) + b = 2$ .

$a (2) + b = 2$

$- 2 = a (- 2) + b$

$- 3 = a (- 3) + b$

Step 1.1.3.1.2

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

$2 a + b = 2$

$- 2 = a (- 2) + b$

$- 3 = a (- 3) + b$

Step 1.1.3.1.3

Subtract $2 a$ from both sides of the equation.

$b = 2 - 2 a$

$- 2 = a (- 2) + b$

$- 3 = a (- 3) + b$

$b = 2 - 2 a$

$- 2 = a (- 2) + b$

$- 3 = a (- 3) + b$

Step 1.1.3.2

Replace all occurrences of $b$ with $2 - 2 a$ in each equation.

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Step 1.1.3.2.1

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

$- 2 = a (- 2) + 2 - 2 a$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

Step 1.1.3.2.2

Simplify $- 2 = a (- 2) + 2 - 2 a$ .

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Step 1.1.3.2.2.1

Simplify the left side.

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Step 1.1.3.2.2.1.1

Remove parentheses.

$- 2 = a (- 2) + 2 - 2 a$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

$- 2 = a (- 2) + 2 - 2 a$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

Step 1.1.3.2.2.2

Simplify the right side.

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Step 1.1.3.2.2.2.1

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

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Step 1.1.3.2.2.2.1.1

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

$- 2 = - 2 a + 2 - 2 a$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

Step 1.1.3.2.2.2.1.2

Subtract $2 a$ from $- 2 a$ .

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = a (- 3) + b$

Step 1.1.3.2.3

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

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

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.2.4

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

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Step 1.1.3.2.4.1

Simplify the left side.

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Step 1.1.3.2.4.1.1

Remove parentheses.

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

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

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

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.2.4.2

Simplify the right side.

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Step 1.1.3.2.4.2.1

Simplify $a (- 3) + 2 - 2 a$ .

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Step 1.1.3.2.4.2.1.1

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

$- 3 = - 3 a + 2 - 2 a$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.2.4.2.1.2

Subtract $2 a$ from $- 3 a$ .

$- 3 = - 5 a + 2$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = - 5 a + 2$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = - 5 a + 2$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = - 5 a + 2$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 3 = - 5 a + 2$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3

Solve for $a$ in $- 3 = - 5 a + 2$ .

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Step 1.1.3.3.1

Rewrite the equation as $- 5 a + 2 = - 3$ .

$- 5 a + 2 = - 3$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3.2

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

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Step 1.1.3.3.2.1

Subtract $2$ from both sides of the equation.

$- 5 a = - 3 - 2$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3.2.2

Subtract $2$ from $- 3$ .

$- 5 a = - 5$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$- 5 a = - 5$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3.3

Divide each term in $- 5 a = - 5$ by $- 5$ and simplify.

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Step 1.1.3.3.3.1

Divide each term in $- 5 a = - 5$ by $- 5$ .

$\frac{- 5 a}{- 5} = \frac{- 5}{- 5}$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3.3.2

Simplify the left side.

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Step 1.1.3.3.3.2.1

Cancel the common factor of $- 5$ .

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Step 1.1.3.3.3.2.1.1

Cancel the common factor.

$\frac{- 5 a}{- 5} = \frac{- 5}{- 5}$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3.3.2.1.2

Divide $a$ by $1$ .

$a = \frac{- 5}{- 5}$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$a = \frac{- 5}{- 5}$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$a = \frac{- 5}{- 5}$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.3.3.3

Simplify the right side.

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Step 1.1.3.3.3.3.1

Divide $- 5$ by $- 5$ .

$a = 1$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$a = 1$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$a = 1$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

$a = 1$

$- 2 = - 4 a + 2$

$b = 2 - 2 a$

Step 1.1.3.4

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

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Step 1.1.3.4.1

Replace all occurrences of $a$ in $- 2 = - 4 a + 2$ with $1$ .

$- 2 = - 4 \cdot 1 + 2$

$a = 1$

$b = 2 - 2 a$

Step 1.1.3.4.2

Simplify the right side.

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Step 1.1.3.4.2.1

Simplify $- 4 \cdot 1 + 2$ .

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Step 1.1.3.4.2.1.1

Multiply $- 4$ by $1$ .

$- 2 = - 4 + 2$

$a = 1$

$b = 2 - 2 a$

Step 1.1.3.4.2.1.2

Add $- 4$ and $2$ .

$- 2 = - 2$

$a = 1$

$b = 2 - 2 a$

$- 2 = - 2$

$a = 1$

$b = 2 - 2 a$

$- 2 = - 2$

$a = 1$

$b = 2 - 2 a$

Step 1.1.3.4.3

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

$b = 2 - 2 \cdot 1$

$- 2 = - 2$

$a = 1$

Step 1.1.3.4.4

Simplify the right side.

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Step 1.1.3.4.4.1

Simplify $2 - 2 \cdot 1$ .

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Step 1.1.3.4.4.1.1

Multiply $- 2$ by $1$ .

$b = 2 - 2$

$- 2 = - 2$

$a = 1$

Step 1.1.3.4.4.1.2

Subtract $2$ from $2$ .

$b = 0$

$- 2 = - 2$

$a = 1$

$b = 0$

$- 2 = - 2$

$a = 1$

$b = 0$

$- 2 = - 2$

$a = 1$

$b = 0$

$- 2 = - 2$

$a = 1$

Step 1.1.3.5

Remove any equations from the system that are always true.

$b = 0$

$a = 1$

Step 1.1.3.6

List all of the solutions.

$b = 0, a = 1$

Step 1.1.4

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 1.1.4.1

Calculate the value of $y$ when $a = 1$ , $b = 0$ , and $x = 2$ .

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Step 1.1.4.1.1

Multiply $2$ by $1$ .

$y = 2 + 0$

Step 1.1.4.1.2

Add $2$ and $0$ .

$y = 2$

Step 1.1.4.2

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 2$ and $y = 2$ .

$2 = 2$

Step 1.1.4.3

Calculate the value of $y$ when $a = 1$ , $b = 0$ , and $x = - 2$ .

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Step 1.1.4.3.1

Multiply $- 2$ by $1$ .

$y = - 2 + 0$

Step 1.1.4.3.2

Add $- 2$ and $0$ .

$y = - 2$

Step 1.1.4.4

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = - 2$ . This check passes since $y = - 2$ and $y = - 2$ .

$- 2 = - 2$

Step 1.1.4.5

Calculate the value of $y$ when $a = 1$ , $b = 0$ , and $x = - 3$ .

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Step 1.1.4.5.1

Multiply $- 3$ by $1$ .

$y = - 3 + 0$

Step 1.1.4.5.2

Add $- 3$ and $0$ .

$y = - 3$

Step 1.1.4.6

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = - 3$ . This check passes since $y = - 3$ and $y = - 3$ .

$- 3 = - 3$

Step 1.1.4.7

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

The function is linear

Step 1.2

Since all $y = y$ , the function is linear and follows the form $y = x$ .

$y = x$

Step 2

Find $x$ .

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Step 2.1

Use the function rule equation to find $x$ .

$x = 1$

Step 2.2

Simplify.

$x = 1$

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