Find the Function Rule table[[x,y],[4,2],[5,3],[6,4]]

$\begin{array}{c} x & y \\ 4 & 2 \\ 5 & 3 \\ 6 & 4 \end{array}$

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 $y = a x + b$ .

$2 = a (4) + b 3 = a (5) + b 4 = a (6) + b$

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

Solve for $b$ in $2 = a (4) + b$ .

Rewrite the equation as $a (4) + b = 2$ .

$a (4) + b = 2$

$3 = a (5) + b$

$4 = a (6) + b$

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

$4 a + b = 2$

$3 = a (5) + b$

$4 = a (6) + b$

Subtract $4 a$ from both sides of the equation.

$b = 2 - 4 a$

$3 = a (5) + b$

$4 = a (6) + b$

$b = 2 - 4 a$

$3 = a (5) + b$

$4 = a (6) + b$

Replace all occurrences of $b$ with $2 - 4 a$ in each equation.

Replace all occurrences of $b$ in $3 = a (5) + b$ with $2 - 4 a$ .

$3 = a (5) + 2 - 4 a$

$b = 2 - 4 a$

$4 = a (6) + b$

Simplify $3 = a (5) + 2 - 4 a$ .

Simplify the left side.

Remove parentheses.

$3 = a (5) + 2 - 4 a$

$b = 2 - 4 a$

$4 = a (6) + b$

$3 = a (5) + 2 - 4 a$

$b = 2 - 4 a$

$4 = a (6) + b$

Simplify the right side.

Simplify $a (5) + 2 - 4 a$ .

Move $5$ to the left of $a$ .

$3 = 5 a + 2 - 4 a$

$b = 2 - 4 a$

$4 = a (6) + b$

Subtract $4 a$ from $5 a$ .

$3 = a + 2$

$b = 2 - 4 a$

$4 = a (6) + b$

$3 = a + 2$

$b = 2 - 4 a$

$4 = a (6) + b$

$3 = a + 2$

$b = 2 - 4 a$

$4 = a (6) + b$

$3 = a + 2$

$b = 2 - 4 a$

$4 = a (6) + b$

Replace all occurrences of $b$ in $4 = a (6) + b$ with $2 - 4 a$ .

$4 = a (6) + 2 - 4 a$

$3 = a + 2$

$b = 2 - 4 a$

Simplify $4 = a (6) + 2 - 4 a$ .

Simplify the left side.

Remove parentheses.

$4 = a (6) + 2 - 4 a$

$3 = a + 2$

$b = 2 - 4 a$

$4 = a (6) + 2 - 4 a$

$3 = a + 2$

$b = 2 - 4 a$

Simplify the right side.

Simplify $a (6) + 2 - 4 a$ .

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

$4 = 6 a + 2 - 4 a$

$3 = a + 2$

$b = 2 - 4 a$

Subtract $4 a$ from $6 a$ .

$4 = 2 a + 2$

$3 = a + 2$

$b = 2 - 4 a$

$4 = 2 a + 2$

$3 = a + 2$

$b = 2 - 4 a$

$4 = 2 a + 2$

$3 = a + 2$

$b = 2 - 4 a$

$4 = 2 a + 2$

$3 = a + 2$

$b = 2 - 4 a$

$4 = 2 a + 2$

$3 = a + 2$

$b = 2 - 4 a$

Solve for $a$ in $4 = 2 a + 2$ .

Rewrite the equation as $2 a + 2 = 4$ .

$2 a + 2 = 4$

$3 = a + 2$

$b = 2 - 4 a$

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

Subtract $2$ from both sides of the equation.

$2 a = 4 - 2$

$3 = a + 2$

$b = 2 - 4 a$

Subtract $2$ from $4$ .

$2 a = 2$

$3 = a + 2$

$b = 2 - 4 a$

$2 a = 2$

$3 = a + 2$

$b = 2 - 4 a$

Divide each term in $2 a = 2$ by $2$ and simplify.

Divide each term in $2 a = 2$ by $2$ .

$\frac{2 a}{2} = \frac{2}{2}$

$3 = a + 2$

$b = 2 - 4 a$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 a}{2} = \frac{2}{2}$

$3 = a + 2$

$b = 2 - 4 a$

Divide $a$ by $1$ .

$a = \frac{2}{2}$

$3 = a + 2$

$b = 2 - 4 a$

$a = \frac{2}{2}$

$3 = a + 2$

$b = 2 - 4 a$

$a = \frac{2}{2}$

$3 = a + 2$

$b = 2 - 4 a$

Simplify the right side.

Divide $2$ by $2$ .

$a = 1$

$3 = a + 2$

$b = 2 - 4 a$

$a = 1$

$3 = a + 2$

$b = 2 - 4 a$

$a = 1$

$3 = a + 2$

$b = 2 - 4 a$

$a = 1$

$3 = a + 2$

$b = 2 - 4 a$

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

Replace all occurrences of $a$ in $3 = a + 2$ with $1$ .

$3 = (1) + 2$

$a = 1$

$b = 2 - 4 a$

Simplify the right side.

Add $1$ and $2$ .

$3 = 3$

$a = 1$

$b = 2 - 4 a$

$3 = 3$

$a = 1$

$b = 2 - 4 a$

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

$b = 2 - 4 \cdot 1$

$3 = 3$

$a = 1$

Simplify the right side.

Simplify $2 - 4 \cdot 1$ .

Multiply $- 4$ by $1$ .

$b = 2 - 4$

$3 = 3$

$a = 1$

Subtract $4$ from $2$ .

$b = - 2$

$3 = 3$

$a = 1$

$b = - 2$

$3 = 3$

$a = 1$

$b = - 2$

$3 = 3$

$a = 1$

$b = - 2$

$3 = 3$

$a = 1$

Remove any equations from the system that are always true.

$b = - 2$

$a = 1$

List all of the solutions.

$b = - 2, a = 1$

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

Multiply $4$ by $1$ .

$y = 4 - 2$

Subtract $2$ from $4$ .

$y = 2$

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

$2 = 2$

Calculate the value of $y$ when $a = 1$ , $b = - 2$ , and $x = 5$ .

Multiply $5$ by $1$ .

$y = 5 - 2$

Subtract $2$ from $5$ .

$y = 3$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 5$ . This check passes since $y = 3$ and $y = 3$ .

$3 = 3$

Calculate the value of $y$ when $a = 1$ , $b = - 2$ , and $x = 6$ .

Multiply $6$ by $1$ .

$y = 6 - 2$

Subtract $2$ from $6$ .

$y = 4$

If the table has a linear function rule, $y = y$ for the corresponding $x$ value, $x = 6$ . This check passes since $y = 4$ and $y = 4$ .

$4 = 4$

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

The function is linear

Since all $y = y$ , the function is linear and follows the form $y = x - 2$ .

$y = x - 2$

Find the Function Rule table[[x,y],[4,2],[5,3],[6,4]]

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