Find the Function Rule table[[x,y],[1,6],[2,10],[3,14]]

$\begin{array}{c} x & y \\ 1 & 6 \\ 2 & 10 \\ 3 & 14 \end{array}$

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 $y = a x + b$ .

$6 = a (1) + b 10 = a (2) + b 14 = a (3) + b$

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

Solve for $a$ in $6 = a + b$ .

Rewrite the equation as $a + b = 6$ .

$a + b = 6$

$10 = a (2) + b$

$14 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 6 - b$

$10 = a (2) + b$

$14 = a (3) + b$

$a = 6 - b$

$10 = a (2) + b$

$14 = a (3) + b$

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

Replace all occurrences of $a$ in $10 = a (2) + b$ with $6 - b$ .

$10 = (6 - b) (2) + b$

$a = 6 - b$

$14 = a (3) + b$

Simplify the right side.

Simplify $(6 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

$10 = 6 \cdot 2 - b \cdot 2 + b$

$a = 6 - b$

$14 = a (3) + b$

Multiply $6$ by $2$ .

$10 = 12 - b \cdot 2 + b$

$a = 6 - b$

$14 = a (3) + b$

Multiply $2$ by $- 1$ .

$10 = 12 - 2 b + b$

$a = 6 - b$

$14 = a (3) + b$

$10 = 12 - 2 b + b$

$a = 6 - b$

$14 = a (3) + b$

Add $- 2 b$ and $b$ .

$10 = 12 - b$

$a = 6 - b$

$14 = a (3) + b$

$10 = 12 - b$

$a = 6 - b$

$14 = a (3) + b$

$10 = 12 - b$

$a = 6 - b$

$14 = a (3) + b$

Replace all occurrences of $a$ in $14 = a (3) + b$ with $6 - b$ .

$14 = (6 - b) (3) + b$

$10 = 12 - b$

$a = 6 - b$

Simplify the right side.

Simplify $(6 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$14 = 6 \cdot 3 - b \cdot 3 + b$

$10 = 12 - b$

$a = 6 - b$

Multiply $6$ by $3$ .

$14 = 18 - b \cdot 3 + b$

$10 = 12 - b$

$a = 6 - b$

Multiply $3$ by $- 1$ .

$14 = 18 - 3 b + b$

$10 = 12 - b$

$a = 6 - b$

$14 = 18 - 3 b + b$

$10 = 12 - b$

$a = 6 - b$

Add $- 3 b$ and $b$ .

$14 = 18 - 2 b$

$10 = 12 - b$

$a = 6 - b$

$14 = 18 - 2 b$

$10 = 12 - b$

$a = 6 - b$

$14 = 18 - 2 b$

$10 = 12 - b$

$a = 6 - b$

$14 = 18 - 2 b$

$10 = 12 - b$

$a = 6 - b$

Solve for $b$ in $14 = 18 - 2 b$ .

Rewrite the equation as $18 - 2 b = 14$ .

$18 - 2 b = 14$

$10 = 12 - b$

$a = 6 - b$

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

Subtract $18$ from both sides of the equation.

$- 2 b = 14 - 18$

$10 = 12 - b$

$a = 6 - b$

Subtract $18$ from $14$ .

$- 2 b = - 4$

$10 = 12 - b$

$a = 6 - b$

$- 2 b = - 4$

$10 = 12 - b$

$a = 6 - b$

Divide each term in $- 2 b = - 4$ by $- 2$ and simplify.

Divide each term in $- 2 b = - 4$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{- 4}{- 2}$

$10 = 12 - b$

$a = 6 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{- 4}{- 2}$

$10 = 12 - b$

$a = 6 - b$

Divide $b$ by $1$ .

$b = \frac{- 4}{- 2}$

$10 = 12 - b$

$a = 6 - b$

$b = \frac{- 4}{- 2}$

$10 = 12 - b$

$a = 6 - b$

$b = \frac{- 4}{- 2}$

$10 = 12 - b$

$a = 6 - b$

Simplify the right side.

Divide $- 4$ by $- 2$ .

$b = 2$

$10 = 12 - b$

$a = 6 - b$

$b = 2$

$10 = 12 - b$

$a = 6 - b$

$b = 2$

$10 = 12 - b$

$a = 6 - b$

$b = 2$

$10 = 12 - b$

$a = 6 - b$

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

Replace all occurrences of $b$ in $10 = 12 - b$ with $2$ .

$10 = 12 - (2)$

$b = 2$

$a = 6 - b$

Simplify the right side.

Simplify $12 - (2)$ .

Multiply $- 1$ by $2$ .

$10 = 12 - 2$

$b = 2$

$a = 6 - b$

Subtract $2$ from $12$ .

$10 = 10$

$b = 2$

$a = 6 - b$

$10 = 10$

$b = 2$

$a = 6 - b$

$10 = 10$

$b = 2$

$a = 6 - b$

Replace all occurrences of $b$ in $a = 6 - b$ with $2$ .

$a = 6 - (2)$

$10 = 10$

$b = 2$

Simplify the right side.

Simplify $6 - (2)$ .

Multiply $- 1$ by $2$ .

$a = 6 - 2$

$10 = 10$

$b = 2$

Subtract $2$ from $6$ .

$a = 4$

$10 = 10$

$b = 2$

$a = 4$

$10 = 10$

$b = 2$

$a = 4$

$10 = 10$

$b = 2$

$a = 4$

$10 = 10$

$b = 2$

Remove any equations from the system that are always true.

$a = 4$

$b = 2$

List all of the solutions.

$a = 4, b = 2$

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 $a = 4$ , $b = 2$ , and $x = 1$ .

Multiply $4$ by $1$ .

$y = 4 + 2$

Add $4$ and $2$ .

$y = 6$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = 6$ and $y = 6$ .

$6 = 6$

Calculate the value of $y$ when $a = 4$ , $b = 2$ , and $x = 2$ .

Multiply $4$ by $2$ .

$y = 8 + 2$

Add $8$ and $2$ .

$y = 10$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 10$ and $y = 10$ .

$10 = 10$

Calculate the value of $y$ when $a = 4$ , $b = 2$ , and $x = 3$ .

Multiply $4$ by $3$ .

$y = 12 + 2$

Add $12$ and $2$ .

$y = 14$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 14$ and $y = 14$ .

$14 = 14$

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 $y = 4 x + 2$ .

$y = 4 x + 2$

Find the Function Rule table[[x,y],[1,6],[2,10],[3,14]]

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