Find the Function Rule table[[x,y],[1,1],[2,4],[3,7]]

$\begin{array}{c} x & y \\ 1 & 1 \\ 2 & 4 \\ 3 & 7 \end{array}$

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 $y = a x + b$ .

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

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

Solve for $a$ in $1 = a + b$ .

Rewrite the equation as $a + b = 1$ .

$a + b = 1$

$4 = a (2) + b$

$7 = a (3) + b$

Subtract $b$ from both sides of the equation.

$a = 1 - b$

$4 = a (2) + b$

$7 = a (3) + b$

$a = 1 - b$

$4 = a (2) + b$

$7 = a (3) + b$

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

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

$4 = (1 - b) (2) + b$

$a = 1 - b$

$7 = a (3) + b$

Simplify the right side.

Simplify $(1 - b) (2) + b$ .

Simplify each term.

Apply the distributive property.

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

$a = 1 - b$

$7 = a (3) + b$

Multiply $2$ by $1$ .

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

$a = 1 - b$

$7 = a (3) + b$

Multiply $2$ by $- 1$ .

$4 = 2 - 2 b + b$

$a = 1 - b$

$7 = a (3) + b$

$4 = 2 - 2 b + b$

$a = 1 - b$

$7 = a (3) + b$

Add $- 2 b$ and $b$ .

$4 = 2 - b$

$a = 1 - b$

$7 = a (3) + b$

$4 = 2 - b$

$a = 1 - b$

$7 = a (3) + b$

$4 = 2 - b$

$a = 1 - b$

$7 = a (3) + b$

Replace all occurrences of $a$ in $7 = a (3) + b$ with $1 - b$ .

$7 = (1 - b) (3) + b$

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Simplify $(1 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$7 = 1 \cdot 3 - b \cdot 3 + b$

$4 = 2 - b$

$a = 1 - b$

Multiply $3$ by $1$ .

$7 = 3 - b \cdot 3 + b$

$4 = 2 - b$

$a = 1 - b$

Multiply $3$ by $- 1$ .

$7 = 3 - 3 b + b$

$4 = 2 - b$

$a = 1 - b$

$7 = 3 - 3 b + b$

$4 = 2 - b$

$a = 1 - b$

Add $- 3 b$ and $b$ .

$7 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$7 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$7 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

$7 = 3 - 2 b$

$4 = 2 - b$

$a = 1 - b$

Solve for $b$ in $7 = 3 - 2 b$ .

Rewrite the equation as $3 - 2 b = 7$ .

$3 - 2 b = 7$

$4 = 2 - b$

$a = 1 - b$

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

Subtract $3$ from both sides of the equation.

$- 2 b = 7 - 3$

$4 = 2 - b$

$a = 1 - b$

Subtract $3$ from $7$ .

$- 2 b = 4$

$4 = 2 - b$

$a = 1 - b$

$- 2 b = 4$

$4 = 2 - b$

$a = 1 - b$

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

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

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

$4 = 2 - b$

$a = 1 - b$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

$4 = 2 - b$

$a = 1 - b$

Divide $b$ by $1$ .

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

$4 = 2 - b$

$a = 1 - b$

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

$4 = 2 - b$

$a = 1 - b$

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

$4 = 2 - b$

$a = 1 - b$

Simplify the right side.

Divide $4$ by $- 2$ .

$b = - 2$

$4 = 2 - b$

$a = 1 - b$

$b = - 2$

$4 = 2 - b$

$a = 1 - b$

$b = - 2$

$4 = 2 - b$

$a = 1 - b$

$b = - 2$

$4 = 2 - b$

$a = 1 - b$

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

Replace all occurrences of $b$ in $4 = 2 - b$ with $- 2$ .

$4 = 2 - (- 2)$

$b = - 2$

$a = 1 - b$

Simplify the right side.

Simplify $2 - (- 2)$ .

Multiply $- 1$ by $- 2$ .

$4 = 2 + 2$

$b = - 2$

$a = 1 - b$

Add $2$ and $2$ .

$4 = 4$

$b = - 2$

$a = 1 - b$

$4 = 4$

$b = - 2$

$a = 1 - b$

$4 = 4$

$b = - 2$

$a = 1 - b$

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

$a = 1 - (- 2)$

$4 = 4$

$b = - 2$

Simplify the right side.

Simplify $1 - (- 2)$ .

Multiply $- 1$ by $- 2$ .

$a = 1 + 2$

$4 = 4$

$b = - 2$

Add $1$ and $2$ .

$a = 3$

$4 = 4$

$b = - 2$

$a = 3$

$4 = 4$

$b = - 2$

$a = 3$

$4 = 4$

$b = - 2$

$a = 3$

$4 = 4$

$b = - 2$

Remove any equations from the system that are always true.

$a = 3$

$b = - 2$

List all of the solutions.

$a = 3, b = - 2$

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 $a = 3$ , $b = - 2$ , and $x = 1$ .

Multiply $3$ by $1$ .

$y = 3 - 2$

Subtract $2$ from $3$ .

$y = 1$

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

$1 = 1$

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

Multiply $3$ by $2$ .

$y = 6 - 2$

Subtract $2$ from $6$ .

$y = 4$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

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

Multiply $3$ by $3$ .

$y = 9 - 2$

Subtract $2$ from $9$ .

$y = 7$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = 7$ and $y = 7$ .

$7 = 7$

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 $y = 3 x - 2$ .

$y = 3 x - 2$

Find the Function Rule table[[x,y],[1,1],[2,4],[3,7]]

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