Find the Function Rule table[[x,y],[2,16.00],[4,32.00],[6,48.00]]

$\begin{array}{c} x & y \\ 2 & 16 \\ 4 & 32 \\ 6 & 48 \end{array}$

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 $y = a x + b$ .

$16 = a (2) + b 32 = a (4) + b 48 = a (6) + b$

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

Solve for $b$ in $16 = a (2) + b$ .

Rewrite the equation as $a (2) + b = 16$ .

$a (2) + b = 16$

$32 = a (4) + b$

$48 = a (6) + b$

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

$2 a + b = 16$

$32 = a (4) + b$

$48 = a (6) + b$

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

$b = 16 - 2 a$

$32 = a (4) + b$

$48 = a (6) + b$

$b = 16 - 2 a$

$32 = a (4) + b$

$48 = a (6) + b$

Replace all occurrences of $b$ with $16 - 2 a$ in each equation.

Replace all occurrences of $b$ in $32 = a (4) + b$ with $16 - 2 a$ .

$32 = a (4) + 16 - 2 a$

$b = 16 - 2 a$

$48 = a (6) + b$

Simplify $32 = a (4) + 16 - 2 a$ .

Simplify the left side.

Remove parentheses.

$32 = a (4) + 16 - 2 a$

$b = 16 - 2 a$

$48 = a (6) + b$

$32 = a (4) + 16 - 2 a$

$b = 16 - 2 a$

$48 = a (6) + b$

Simplify the right side.

Simplify $a (4) + 16 - 2 a$ .

Move $4$ to the left of $a$ .

$32 = 4 a + 16 - 2 a$

$b = 16 - 2 a$

$48 = a (6) + b$

Subtract $2 a$ from $4 a$ .

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = a (6) + b$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = a (6) + b$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = a (6) + b$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = a (6) + b$

Replace all occurrences of $b$ in $48 = a (6) + b$ with $16 - 2 a$ .

$48 = a (6) + 16 - 2 a$

$32 = 2 a + 16$

$b = 16 - 2 a$

Simplify $48 = a (6) + 16 - 2 a$ .

Simplify the left side.

Remove parentheses.

$48 = a (6) + 16 - 2 a$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = a (6) + 16 - 2 a$

$32 = 2 a + 16$

$b = 16 - 2 a$

Simplify the right side.

Simplify $a (6) + 16 - 2 a$ .

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

$48 = 6 a + 16 - 2 a$

$32 = 2 a + 16$

$b = 16 - 2 a$

Subtract $2 a$ from $6 a$ .

$48 = 4 a + 16$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = 4 a + 16$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = 4 a + 16$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = 4 a + 16$

$32 = 2 a + 16$

$b = 16 - 2 a$

$48 = 4 a + 16$

$32 = 2 a + 16$

$b = 16 - 2 a$

Solve for $a$ in $48 = 4 a + 16$ .

Rewrite the equation as $4 a + 16 = 48$ .

$4 a + 16 = 48$

$32 = 2 a + 16$

$b = 16 - 2 a$

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

Subtract $16$ from both sides of the equation.

$4 a = 48 - 16$

$32 = 2 a + 16$

$b = 16 - 2 a$

Subtract $16$ from $48$ .

$4 a = 32$

$32 = 2 a + 16$

$b = 16 - 2 a$

$4 a = 32$

$32 = 2 a + 16$

$b = 16 - 2 a$

Divide each term in $4 a = 32$ by $4$ and simplify.

Divide each term in $4 a = 32$ by $4$ .

$\frac{4 a}{4} = \frac{32}{4}$

$32 = 2 a + 16$

$b = 16 - 2 a$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 a}{4} = \frac{32}{4}$

$32 = 2 a + 16$

$b = 16 - 2 a$

Divide $a$ by $1$ .

$a = \frac{32}{4}$

$32 = 2 a + 16$

$b = 16 - 2 a$

$a = \frac{32}{4}$

$32 = 2 a + 16$

$b = 16 - 2 a$

$a = \frac{32}{4}$

$32 = 2 a + 16$

$b = 16 - 2 a$

Simplify the right side.

Divide $32$ by $4$ .

$a = 8$

$32 = 2 a + 16$

$b = 16 - 2 a$

$a = 8$

$32 = 2 a + 16$

$b = 16 - 2 a$

$a = 8$

$32 = 2 a + 16$

$b = 16 - 2 a$

$a = 8$

$32 = 2 a + 16$

$b = 16 - 2 a$

Replace all occurrences of $a$ with $8$ in each equation.

Replace all occurrences of $a$ in $32 = 2 a + 16$ with $8$ .

$32 = 2 (8) + 16$

$a = 8$

$b = 16 - 2 a$

Simplify the right side.

Simplify $2 (8) + 16$ .

Multiply $2$ by $8$ .

$32 = 16 + 16$

$a = 8$

$b = 16 - 2 a$

Add $16$ and $16$ .

$32 = 32$

$a = 8$

$b = 16 - 2 a$

$32 = 32$

$a = 8$

$b = 16 - 2 a$

$32 = 32$

$a = 8$

$b = 16 - 2 a$

Replace all occurrences of $a$ in $b = 16 - 2 a$ with $8$ .

$b = 16 - 2 \cdot 8$

$32 = 32$

$a = 8$

Simplify the right side.

Simplify $16 - 2 \cdot 8$ .

Multiply $- 2$ by $8$ .

$b = 16 - 16$

$32 = 32$

$a = 8$

Subtract $16$ from $16$ .

$b = 0$

$32 = 32$

$a = 8$

$b = 0$

$32 = 32$

$a = 8$

$b = 0$

$32 = 32$

$a = 8$

$b = 0$

$32 = 32$

$a = 8$

Remove any equations from the system that are always true.

$b = 0$

$a = 8$

List all of the solutions.

$b = 0, a = 8$

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 $a = 8$ , $b = 0$ , and $x = 2$ .

Multiply $8$ by $2$ .

$y = 16 + 0$

Add $16$ and $0$ .

$y = 16$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 16$ and $y = 16$ .

$16 = 16$

Calculate the value of $y$ when $a = 8$ , $b = 0$ , and $x = 4$ .

Multiply $8$ by $4$ .

$y = 32 + 0$

Add $32$ and $0$ .

$y = 32$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 32$ and $y = 32$ .

$32 = 32$

Calculate the value of $y$ when $a = 8$ , $b = 0$ , and $x = 6$ .

Multiply $8$ by $6$ .

$y = 48 + 0$

Add $48$ and $0$ .

$y = 48$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 6$ . This check passes since $y = 48$ and $y = 48$ .

$48 = 48$

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 $y = 8 x$ .

$y = 8 x$

Find the Function Rule table[[x,y],[2,16.00],[4,32.00],[6,48.00]]

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