Find the Function Rule table[[x,y],[1,34],[2,50],[3,66]]

$\begin{array}{c} x & y \\ 1 & 34 \\ 2 & 50 \\ 3 & 66 \end{array}$

Step 1

Check if the function rule is linear.

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

$y = a x + b$

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

$34 = a (1) + b 50 = a (2) + b 66 = a (3) + b$

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

Solve for $a$ in $34 = a + b$ .

Rewrite the equation as $a + b = 34$ .

$a + b = 34$

$50 = a (2) + b$

$66 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 34 - b$

$50 = a (2) + b$

$66 = a (3) + b$

$a = 34 - b$

$50 = a (2) + b$

$66 = a (3) + b$

Replace all occurrences of $a$ with $34 - b$ in each equation.

Replace all occurrences of $a$ in $50 = a (2) + b$ with $34 - b$ .

$50 = (34 - b) (2) + b$

$a = 34 - b$

$66 = a (3) + b$

Simplify the right side.

Simplify $(34 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

$50 = 34 \cdot 2 - b \cdot 2 + b$

$a = 34 - b$

$66 = a (3) + b$

Multiply $34$ by $2$ .

$50 = 68 - b \cdot 2 + b$

$a = 34 - b$

$66 = a (3) + b$

Multiply $2$ by $- 1$ .

$50 = 68 - 2 b + b$

$a = 34 - b$

$66 = a (3) + b$

$50 = 68 - 2 b + b$

$a = 34 - b$

$66 = a (3) + b$

Add $- 2 b$ and $b$ .

$50 = 68 - b$

$a = 34 - b$

$66 = a (3) + b$

$50 = 68 - b$

$a = 34 - b$

$66 = a (3) + b$

$50 = 68 - b$

$a = 34 - b$

$66 = a (3) + b$

Replace all occurrences of $a$ in $66 = a (3) + b$ with $34 - b$ .

$66 = (34 - b) (3) + b$

$50 = 68 - b$

$a = 34 - b$

Simplify the right side.

Simplify $(34 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$66 = 34 \cdot 3 - b \cdot 3 + b$

$50 = 68 - b$

$a = 34 - b$

Multiply $34$ by $3$ .

$66 = 102 - b \cdot 3 + b$

$50 = 68 - b$

$a = 34 - b$

Multiply $3$ by $- 1$ .

$66 = 102 - 3 b + b$

$50 = 68 - b$

$a = 34 - b$

$66 = 102 - 3 b + b$

$50 = 68 - b$

$a = 34 - b$

Add $- 3 b$ and $b$ .

$66 = 102 - 2 b$

$50 = 68 - b$

$a = 34 - b$

$66 = 102 - 2 b$

$50 = 68 - b$

$a = 34 - b$

$66 = 102 - 2 b$

$50 = 68 - b$

$a = 34 - b$

$66 = 102 - 2 b$

$50 = 68 - b$

$a = 34 - b$

Solve for $b$ in $66 = 102 - 2 b$ .

Rewrite the equation as $102 - 2 b = 66$ .

$102 - 2 b = 66$

$50 = 68 - b$

$a = 34 - b$

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

Subtract $102$ from both sides of the equation.

$- 2 b = 66 - 102$

$50 = 68 - b$

$a = 34 - b$

Subtract $102$ from $66$ .

$- 2 b = - 36$

$50 = 68 - b$

$a = 34 - b$

$- 2 b = - 36$

$50 = 68 - b$

$a = 34 - b$

Divide each term in $- 2 b = - 36$ by $- 2$ and simplify.

Divide each term in $- 2 b = - 36$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{- 36}{- 2}$

$50 = 68 - b$

$a = 34 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{- 36}{- 2}$

$50 = 68 - b$

$a = 34 - b$

Divide $b$ by $1$ .

$b = \frac{- 36}{- 2}$

$50 = 68 - b$

$a = 34 - b$

$b = \frac{- 36}{- 2}$

$50 = 68 - b$

$a = 34 - b$

$b = \frac{- 36}{- 2}$

$50 = 68 - b$

$a = 34 - b$

Simplify the right side.

Divide $- 36$ by $- 2$ .

$b = 18$

$50 = 68 - b$

$a = 34 - b$

$b = 18$

$50 = 68 - b$

$a = 34 - b$

$b = 18$

$50 = 68 - b$

$a = 34 - b$

$b = 18$

$50 = 68 - b$

$a = 34 - b$

Replace all occurrences of $b$ with $18$ in each equation.

Replace all occurrences of $b$ in $50 = 68 - b$ with $18$ .

$50 = 68 - (18)$

$b = 18$

$a = 34 - b$

Simplify the right side.

Simplify $68 - (18)$ .

Multiply $- 1$ by $18$ .

$50 = 68 - 18$

$b = 18$

$a = 34 - b$

Subtract $18$ from $68$ .

$50 = 50$

$b = 18$

$a = 34 - b$

$50 = 50$

$b = 18$

$a = 34 - b$

$50 = 50$

$b = 18$

$a = 34 - b$

Replace all occurrences of $b$ in $a = 34 - b$ with $18$ .

$a = 34 - (18)$

$50 = 50$

$b = 18$

Simplify the right side.

Simplify $34 - (18)$ .

Multiply $- 1$ by $18$ .

$a = 34 - 18$

$50 = 50$

$b = 18$

Subtract $18$ from $34$ .

$a = 16$

$50 = 50$

$b = 18$

$a = 16$

$50 = 50$

$b = 18$

$a = 16$

$50 = 50$

$b = 18$

$a = 16$

$50 = 50$

$b = 18$

Remove any equations from the system that are always true.

$a = 16$

$b = 18$

List all of the solutions.

$a = 16, b = 18$

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

Calculate the value of $y$ when $a = 16$ , $b = 18$ , and $x = 1$ .

Multiply $16$ by $1$ .

$y = 16 + 18$

Add $16$ and $18$ .

$y = 34$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 34$ and $y = 34$ .

$34 = 34$

Calculate the value of $y$ when $a = 16$ , $b = 18$ , and $x = 2$ .

Multiply $16$ by $2$ .

$y = 32 + 18$

Add $32$ and $18$ .

$y = 50$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 50$ and $y = 50$ .

$50 = 50$

Calculate the value of $y$ when $a = 16$ , $b = 18$ , and $x = 3$ .

Multiply $16$ by $3$ .

$y = 48 + 18$

Add $48$ and $18$ .

$y = 66$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 66$ and $y = 66$ .

$66 = 66$

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

The function is linear

Step 2

Since all $y = y$ , the function is linear and follows the form $y = 16 x + 18$ .

$y = 16 x + 18$

Find the Function Rule table[[x,y],[1,34],[2,50],[3,66]]

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