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

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

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 $y = a x + b$ .

$- 4 = a (1) + b - 1 = a (3) + b 3 = a (4) + b$

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

Solve for $a$ in $- 4 = a + b$ .

Rewrite the equation as $a + b = - 4$ .

$a + b = - 4$

$- 1 = a (3) + b$

$3 = a (4) + b$

Subtract $b$ from both sides of the equation.

$a = - 4 - b$

$- 1 = a (3) + b$

$3 = a (4) + b$

$a = - 4 - b$

$- 1 = a (3) + b$

$3 = a (4) + b$

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

Replace all occurrences of $a$ in $- 1 = a (3) + b$ with $- 4 - b$ .

$- 1 = (- 4 - b) (3) + b$

$a = - 4 - b$

$3 = a (4) + b$

Simplify the right side.

Simplify $(- 4 - b) (3) + b$ .

Simplify each term.

Apply the distributive property.

$- 1 = - 4 \cdot 3 - b \cdot 3 + b$

$a = - 4 - b$

$3 = a (4) + b$

Multiply $- 4$ by $3$ .

$- 1 = - 12 - b \cdot 3 + b$

$a = - 4 - b$

$3 = a (4) + b$

Multiply $3$ by $- 1$ .

$- 1 = - 12 - 3 b + b$

$a = - 4 - b$

$3 = a (4) + b$

$- 1 = - 12 - 3 b + b$

$a = - 4 - b$

$3 = a (4) + b$

Add $- 3 b$ and $b$ .

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = a (4) + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = a (4) + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = a (4) + b$

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

$3 = (- 4 - b) (4) + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Simplify the right side.

Simplify $(- 4 - b) (4) + b$ .

Simplify each term.

Apply the distributive property.

$3 = - 4 \cdot 4 - b \cdot 4 + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Multiply $- 4$ by $4$ .

$3 = - 16 - b \cdot 4 + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Multiply $4$ by $- 1$ .

$3 = - 16 - 4 b + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = - 16 - 4 b + b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Add $- 4 b$ and $b$ .

$3 = - 16 - 3 b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = - 16 - 3 b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = - 16 - 3 b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$3 = - 16 - 3 b$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Solve for $b$ in $3 = - 16 - 3 b$ .

Rewrite the equation as $- 16 - 3 b = 3$ .

$- 16 - 3 b = 3$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

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

Add $16$ to both sides of the equation.

$- 3 b = 3 + 16$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Add $3$ and $16$ .

$- 3 b = 19$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$- 3 b = 19$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Divide each term in $- 3 b = 19$ by $- 3$ and simplify.

Divide each term in $- 3 b = 19$ by $- 3$ .

$\frac{- 3 b}{- 3} = \frac{19}{- 3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 b}{- 3} = \frac{19}{- 3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Divide $b$ by $1$ .

$b = \frac{19}{- 3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$b = \frac{19}{- 3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$b = \frac{19}{- 3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Simplify the right side.

Move the negative in front of the fraction.

$b = - \frac{19}{3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$b = - \frac{19}{3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$b = - \frac{19}{3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

$b = - \frac{19}{3}$

$- 1 = - 12 - 2 b$

$a = - 4 - b$

Replace all occurrences of $b$ with $- \frac{19}{3}$ in each equation.

Replace all occurrences of $b$ in $- 1 = - 12 - 2 b$ with $- \frac{19}{3}$ .

$- 1 = - 12 - 2 (- \frac{19}{3})$

$b = - \frac{19}{3}$

$a = - 4 - b$

Simplify the right side.

Simplify $- 12 - 2 (- \frac{19}{3})$ .

Multiply $- 2 (- \frac{19}{3})$ .

Multiply $- 1$ by $- 2$ .

$- 1 = - 12 + 2 (\frac{19}{3})$

$b = - \frac{19}{3}$

$a = - 4 - b$

Combine $2$ and $\frac{19}{3}$ .

$- 1 = - 12 + \frac{2 \cdot 19}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

Multiply $2$ by $19$ .

$- 1 = - 12 + \frac{38}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

$- 1 = - 12 + \frac{38}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

To write $- 12$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 1 = - 12 \cdot \frac{3}{3} + \frac{38}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

Combine $- 12$ and $\frac{3}{3}$ .

$- 1 = \frac{- 12 \cdot 3}{3} + \frac{38}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

Combine the numerators over the common denominator.

$- 1 = \frac{- 12 \cdot 3 + 38}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

Simplify the numerator.

Multiply $- 12$ by $3$ .

$- 1 = \frac{- 36 + 38}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

Add $- 36$ and $38$ .

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = - 4 - b$

Replace all occurrences of $b$ in $a = - 4 - b$ with $- \frac{19}{3}$ .

$a = - 4 - (- \frac{19}{3})$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Simplify the right side.

Simplify $- 4 - (- \frac{19}{3})$ .

Multiply $- (- \frac{19}{3})$ .

Multiply $- 1$ by $- 1$ .

$a = - 4 + 1 (\frac{19}{3})$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Multiply $\frac{19}{3}$ by $1$ .

$a = - 4 + \frac{19}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = - 4 + \frac{19}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = - 4 \cdot \frac{3}{3} + \frac{19}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Combine $- 4$ and $\frac{3}{3}$ .

$a = \frac{- 4 \cdot 3}{3} + \frac{19}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Combine the numerators over the common denominator.

$a = \frac{- 4 \cdot 3 + 19}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Simplify the numerator.

Multiply $- 4$ by $3$ .

$a = \frac{- 12 + 19}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Add $- 12$ and $19$ .

$a = \frac{7}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = \frac{7}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = \frac{7}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = \frac{7}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

$a = \frac{7}{3}$

$- 1 = \frac{2}{3}$

$b = - \frac{19}{3}$

Since $- 1 = \frac{2}{3}$ is not true, there is no solution.

No solution

Since $y \neq y$ for the corresponding $x$ values, the function is not linear.

The function is not linear

Step 2

Check if the function rule is quadratic.

To find if the table follows a function rule, check whether the function rule could follow the form $y = a x^{2} + b x + c$ .

$y = a x^{2} + b x + c$

Build a set of $3$ equations from the table such that $y = a x^{2} + b x + c$ .

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

Solve for $a$ in $- 4 = a + b + c$ .

Rewrite the equation as $a + b + c = - 4$ .

$a + b + c = - 4$

$- 1 = a \cdot 3^{2} + b (3) + c$

$3 = a \cdot 4^{2} + b (4) + c$

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

Subtract $b$ from both sides of the equation.

$a + c = - 4 - b$

$- 1 = a \cdot 3^{2} + b (3) + c$

$3 = a \cdot 4^{2} + b (4) + c$

Subtract $c$ from both sides of the equation.

$a = - 4 - b - c$

$- 1 = a \cdot 3^{2} + b (3) + c$

$3 = a \cdot 4^{2} + b (4) + c$

$a = - 4 - b - c$

$- 1 = a \cdot 3^{2} + b (3) + c$

$3 = a \cdot 4^{2} + b (4) + c$

$a = - 4 - b - c$

$- 1 = a \cdot 3^{2} + b (3) + c$

$3 = a \cdot 4^{2} + b (4) + c$

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

Replace all occurrences of $a$ in $- 1 = a \cdot 3^{2} + b (3) + c$ with $- 4 - b - c$ .

$- 1 = (- 4 - b - c) \cdot 3^{2} + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Simplify the right side.

Simplify $(- 4 - b - c) \cdot 3^{2} + b (3) + c$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$- 1 = (- 4 - b - c) \cdot 9 + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Apply the distributive property.

$- 1 = - 4 \cdot 9 - b \cdot 9 - c \cdot 9 + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Simplify.

Multiply $- 4$ by $9$ .

$- 1 = - 36 - b \cdot 9 - c \cdot 9 + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Multiply $9$ by $- 1$ .

$- 1 = - 36 - 9 b - c \cdot 9 + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Multiply $9$ by $- 1$ .

$- 1 = - 36 - 9 b - 9 c + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

$- 1 = - 36 - 9 b - 9 c + b (3) + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Move $3$ to the left of $b$ .

$- 1 = - 36 - 9 b - 9 c + 3 b + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

$- 1 = - 36 - 9 b - 9 c + 3 b + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Simplify by adding terms.

Add $- 9 b$ and $3 b$ .

$- 1 = - 36 - 6 b - 9 c + c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Add $- 9 c$ and $c$ .

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = a \cdot 4^{2} + b (4) + c$

Replace all occurrences of $a$ in $3 = a \cdot 4^{2} + b (4) + c$ with $- 4 - b - c$ .

$3 = (- 4 - b - c) \cdot 4^{2} + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Simplify the right side.

Simplify $(- 4 - b - c) \cdot 4^{2} + b (4) + c$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$3 = (- 4 - b - c) \cdot 16 + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Apply the distributive property.

$3 = - 4 \cdot 16 - b \cdot 16 - c \cdot 16 + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Simplify.

Multiply $- 4$ by $16$ .

$3 = - 64 - b \cdot 16 - c \cdot 16 + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Multiply $16$ by $- 1$ .

$3 = - 64 - 16 b - c \cdot 16 + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Multiply $16$ by $- 1$ .

$3 = - 64 - 16 b - 16 c + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = - 64 - 16 b - 16 c + b (4) + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

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

$3 = - 64 - 16 b - 16 c + 4 b + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = - 64 - 16 b - 16 c + 4 b + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Simplify by adding terms.

Add $- 16 b$ and $4 b$ .

$3 = - 64 - 12 b - 16 c + c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Add $- 16 c$ and $c$ .

$3 = - 64 - 12 b - 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = - 64 - 12 b - 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = - 64 - 12 b - 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = - 64 - 12 b - 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$3 = - 64 - 12 b - 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Solve for $b$ in $3 = - 64 - 12 b - 15 c$ .

Rewrite the equation as $- 64 - 12 b - 15 c = 3$ .

$- 64 - 12 b - 15 c = 3$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

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

Add $64$ to both sides of the equation.

$- 12 b - 15 c = 3 + 64$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Add $15 c$ to both sides of the equation.

$- 12 b = 3 + 64 + 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Add $3$ and $64$ .

$- 12 b = 67 + 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$- 12 b = 67 + 15 c$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Divide each term in $- 12 b = 67 + 15 c$ by $- 12$ and simplify.

Divide each term in $- 12 b = 67 + 15 c$ by $- 12$ .

$\frac{- 12 b}{- 12} = \frac{67}{- 12} + \frac{15 c}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 b}{- 12} = \frac{67}{- 12} + \frac{15 c}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Divide $b$ by $1$ .

$b = \frac{67}{- 12} + \frac{15 c}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = \frac{67}{- 12} + \frac{15 c}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = \frac{67}{- 12} + \frac{15 c}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$b = - \frac{67}{12} + \frac{15 c}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Cancel the common factor of $15$ and $- 12$ .

Factor $3$ out of $15 c$ .

$b = - \frac{67}{12} + \frac{3 (5 c)}{- 12}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Cancel the common factors.

Factor $3$ out of $- 12$ .

$b = - \frac{67}{12} + \frac{3 (5 c)}{3 (- 4)}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Cancel the common factor.

$b = - \frac{67}{12} + \frac{3 (5 c)}{3 \cdot - 4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Rewrite the expression.

$b = - \frac{67}{12} + \frac{5 c}{- 4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = - \frac{67}{12} + \frac{5 c}{- 4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = - \frac{67}{12} + \frac{5 c}{- 4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Move the negative in front of the fraction.

$b = - \frac{67}{12} - \frac{5 c}{4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$- 1 = - 36 - 6 b - 8 c$

$a = - 4 - b - c$

Replace all occurrences of $b$ with $- \frac{67}{12} - \frac{5 c}{4}$ in each equation.

Replace all occurrences of $b$ in $- 1 = - 36 - 6 b - 8 c$ with $- \frac{67}{12} - \frac{5 c}{4}$ .

$- 1 = - 36 - 6 (- \frac{67}{12} - \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Simplify the right side.

Simplify $- 36 - 6 (- \frac{67}{12} - \frac{5 c}{4}) - 8 c$ .

Simplify each term.

Apply the distributive property.

$- 1 = - 36 - 6 (- \frac{67}{12}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{67}{12}$ into the numerator.

$- 1 = - 36 - 6 (\frac{- 67}{12}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Factor $6$ out of $- 6$ .

$- 1 = - 36 + 6 (- 1) (\frac{- 67}{12}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Factor $6$ out of $12$ .

$- 1 = - 36 + 6 \cdot (- 1 \frac{- 67}{6 \cdot 2}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Cancel the common factor.

$- 1 = - 36 + 6 \cdot (- 1 \frac{- 67}{6 \cdot 2}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Rewrite the expression.

$- 1 = - 36 - 1 (\frac{- 67}{2}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - 36 - 1 (\frac{- 67}{2}) - 6 (- \frac{5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5 c}{4}$ into the numerator.

$- 1 = - 36 - 1 (\frac{- 67}{2}) - 6 \frac{- 5 c}{4} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Factor $2$ out of $- 6$ .

$- 1 = - 36 - 1 (\frac{- 67}{2}) + 2 (- 3) (\frac{- 5 c}{4}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Factor $2$ out of $4$ .

$- 1 = - 36 - 1 (\frac{- 67}{2}) + 2 \cdot (- 3 \frac{- 5 c}{2 \cdot 2}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Cancel the common factor.

$- 1 = - 36 - 1 (\frac{- 67}{2}) + 2 \cdot (- 3 \frac{- 5 c}{2 \cdot 2}) - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Rewrite the expression.

$- 1 = - 36 - 1 (\frac{- 67}{2}) - 3 \frac{- 5 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - 36 - 1 (\frac{- 67}{2}) - 3 \frac{- 5 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Combine $- 3$ and $\frac{- 5 c}{2}$ .

$- 1 = - 36 - 1 (\frac{- 67}{2}) + \frac{- 3 (- 5 c)}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Multiply $- 5$ by $- 3$ .

$- 1 = - 36 - 1 (\frac{- 67}{2}) + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Simplify each term.

Move the negative in front of the fraction.

$- 1 = - 36 - 1 (- \frac{67}{2}) + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Multiply $- 1 (- \frac{67}{2})$ .

Multiply $- 1$ by $- 1$ .

$- 1 = - 36 + 1 (\frac{67}{2}) + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Multiply $\frac{67}{2}$ by $1$ .

$- 1 = - 36 + \frac{67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - 36 + \frac{67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - 36 + \frac{67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - 36 + \frac{67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

To write $- 36$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 1 = - 36 \cdot \frac{2}{2} + \frac{67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Combine $- 36$ and $\frac{2}{2}$ .

$- 1 = \frac{- 36 \cdot 2}{2} + \frac{67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Combine the numerators over the common denominator.

$- 1 = \frac{- 36 \cdot 2 + 67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Simplify the numerator.

Multiply $- 36$ by $2$ .

$- 1 = \frac{- 72 + 67}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Add $- 72$ and $67$ .

$- 1 = \frac{- 5}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = \frac{- 5}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Move the negative in front of the fraction.

$- 1 = - \frac{5}{2} + \frac{15 c}{2} - 8 c$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

To write $- 8 c$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- 1 = - \frac{5}{2} + \frac{15 c}{2} - 8 c \cdot \frac{2}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Combine $- 8 c$ and $\frac{2}{2}$ .

$- 1 = - \frac{5}{2} + \frac{15 c}{2} + \frac{- 8 c \cdot 2}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Combine the numerators over the common denominator.

$- 1 = - \frac{5}{2} + \frac{15 c - 8 c \cdot 2}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Combine the numerators over the common denominator.

$- 1 = \frac{- 5 + 15 c - 8 c \cdot 2}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Multiply $2$ by $- 8$ .

$- 1 = \frac{- 5 + 15 c - 16 c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Subtract $16 c$ from $15 c$ .

$- 1 = \frac{- 5 - c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Rewrite $- 5$ as $- 1 (5)$ .

$- 1 = \frac{- 1 \cdot 5 - c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Factor $- 1$ out of $- c$ .

$- 1 = \frac{- 1 \cdot 5 - (c)}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Factor $- 1$ out of $- 1 (5) - (c)$ .

$- 1 = \frac{- 1 (5 + c)}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Move the negative in front of the fraction.

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 - b - c$

Replace all occurrences of $b$ in $a = - 4 - b - c$ with $- \frac{67}{12} - \frac{5 c}{4}$ .

$a = - 4 - (- \frac{67}{12} - \frac{5 c}{4}) - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify the right side.

Simplify $- 4 - (- \frac{67}{12} - \frac{5 c}{4}) - c$ .

Simplify each term.

Apply the distributive property.

$a = - 4 + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $- - \frac{67}{12}$ .

Multiply $- 1$ by $- 1$ .

$a = - 4 + 1 (\frac{67}{12}) + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $\frac{67}{12}$ by $1$ .

$a = - 4 + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $- - \frac{5 c}{4}$ .

Multiply $- 1$ by $- 1$ .

$a = - 4 + \frac{67}{12} + 1 (\frac{5 c}{4}) - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $\frac{5 c}{4}$ by $1$ .

$a = - 4 + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = - 4 + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$a = - 4 \cdot \frac{12}{12} + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Combine $- 4$ and $\frac{12}{12}$ .

$a = \frac{- 4 \cdot 12}{12} + \frac{67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Combine the numerators over the common denominator.

$a = \frac{- 4 \cdot 12 + 67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify the numerator.

Multiply $- 4$ by $12$ .

$a = \frac{- 48 + 67}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Add $- 48$ and $67$ .

$a = \frac{19}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{5 c}{4} - c$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

To write $- c$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$a = \frac{19}{12} + \frac{5 c}{4} - c \cdot \frac{4}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify terms.

Combine $- c$ and $\frac{4}{4}$ .

$a = \frac{19}{12} + \frac{5 c}{4} + \frac{- c \cdot 4}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Combine the numerators over the common denominator.

$a = \frac{19}{12} + \frac{5 c - c \cdot 4}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{5 c - c \cdot 4}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $5 c - c \cdot 4$ .

Factor $c$ out of $5 c$ .

$a = \frac{19}{12} + \frac{c \cdot 5 - c \cdot 4}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Factor $c$ out of $- c \cdot 4$ .

$a = \frac{19}{12} + \frac{c \cdot 5 + c (- 1 \cdot 4)}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Factor $c$ out of $c \cdot 5 + c (- 1 \cdot 4)$ .

$a = \frac{19}{12} + \frac{c (5 - 1 \cdot 4)}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{c (5 - 1 \cdot 4)}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $- 1$ by $4$ .

$a = \frac{19}{12} + \frac{c (5 - 4)}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Subtract $4$ from $5$ .

$a = \frac{19}{12} + \frac{c \cdot 1}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{c \cdot 1}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $c$ by $1$ .

$a = \frac{19}{12} + \frac{c}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{c}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{c}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{c}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} + \frac{c}{4}$

$- 1 = - \frac{5 + c}{2}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Solve for $c$ in $- 1 = - \frac{5 + c}{2}$ .

Rewrite the equation as $- \frac{5 + c}{2} = - 1$ .

$- \frac{5 + c}{2} = - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply both sides of the equation by $- 2$ .

$- 2 (- \frac{5 + c}{2}) = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- 2 (- \frac{5 + c}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5 + c}{2}$ into the numerator.

$- 2 \frac{- (5 + c)}{2} = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Factor $2$ out of $- 2$ .

$2 (- 1) (\frac{- (5 + c)}{2}) = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Cancel the common factor.

$2 \cdot (- 1 \frac{- (5 + c)}{2}) = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Rewrite the expression.

$5 + c = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$5 + c = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 (5 + c) = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $5 + c$ by $1$ .

$5 + c = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$5 + c = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$5 + c = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$5 + c = - 2 \cdot - 1$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify the right side.

Multiply $- 2$ by $- 1$ .

$5 + c = 2$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$5 + c = 2$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$5 + c = 2$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

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

Subtract $5$ from both sides of the equation.

$c = 2 - 5$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Subtract $5$ from $2$ .

$c = - 3$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$c = - 3$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$c = - 3$

$a = \frac{19}{12} + \frac{c}{4}$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Replace all occurrences of $c$ with $- 3$ in each equation.

Replace all occurrences of $c$ in $a = \frac{19}{12} + \frac{c}{4}$ with $- 3$ .

$a = \frac{19}{12} + \frac{- 3}{4}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify the right side.

Simplify $\frac{19}{12} + \frac{- 3}{4}$ .

Move the negative in front of the fraction.

$a = \frac{19}{12} - \frac{3}{4}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

To write $- \frac{3}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = \frac{19}{12} - \frac{3}{4} \cdot \frac{3}{3}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{4}$ by $\frac{3}{3}$ .

$a = \frac{19}{12} - \frac{3 \cdot 3}{4 \cdot 3}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Multiply $4$ by $3$ .

$a = \frac{19}{12} - \frac{3 \cdot 3}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{19}{12} - \frac{3 \cdot 3}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Combine the numerators over the common denominator.

$a = \frac{19 - 3 \cdot 3}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Simplify the numerator.

Multiply $- 3$ by $3$ .

$a = \frac{19 - 9}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Subtract $9$ from $19$ .

$a = \frac{10}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{10}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Cancel the common factor of $10$ and $12$ .

Factor $2$ out of $10$ .

$a = \frac{2 (5)}{12}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Cancel the common factors.

Factor $2$ out of $12$ .

$a = \frac{2 \cdot 5}{2 \cdot 6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Cancel the common factor.

$a = \frac{2 \cdot 5}{2 \cdot 6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Rewrite the expression.

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} - \frac{5 c}{4}$

Replace all occurrences of $c$ in $b = - \frac{67}{12} - \frac{5 c}{4}$ with $- 3$ .

$b = - \frac{67}{12} - \frac{5 (- 3)}{4}$

$a = \frac{5}{6}$

$c = - 3$

Simplify the right side.

Simplify $- \frac{67}{12} - \frac{5 (- 3)}{4}$ .

Simplify each term.

Multiply $5$ by $- 3$ .

$b = - \frac{67}{12} - \frac{- 15}{4}$

$a = \frac{5}{6}$

$c = - 3$

Move the negative in front of the fraction.

$b = - \frac{67}{12} + \frac{15}{4}$

$a = \frac{5}{6}$

$c = - 3$

Multiply $- - \frac{15}{4}$ .

Multiply $- 1$ by $- 1$ .

$b = - \frac{67}{12} + 1 (\frac{15}{4})$

$a = \frac{5}{6}$

$c = - 3$

Multiply $\frac{15}{4}$ by $1$ .

$b = - \frac{67}{12} + \frac{15}{4}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} + \frac{15}{4}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} + \frac{15}{4}$

$a = \frac{5}{6}$

$c = - 3$

To write $\frac{15}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$b = - \frac{67}{12} + \frac{15}{4} \cdot \frac{3}{3}$

$a = \frac{5}{6}$

$c = - 3$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{15}{4}$ by $\frac{3}{3}$ .

$b = - \frac{67}{12} + \frac{15 \cdot 3}{4 \cdot 3}$

$a = \frac{5}{6}$

$c = - 3$

Multiply $4$ by $3$ .

$b = - \frac{67}{12} + \frac{15 \cdot 3}{12}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{67}{12} + \frac{15 \cdot 3}{12}$

$a = \frac{5}{6}$

$c = - 3$

Combine the numerators over the common denominator.

$b = \frac{- 67 + 15 \cdot 3}{12}$

$a = \frac{5}{6}$

$c = - 3$

Simplify the numerator.

Multiply $15$ by $3$ .

$b = \frac{- 67 + 45}{12}$

$a = \frac{5}{6}$

$c = - 3$

Add $- 67$ and $45$ .

$b = \frac{- 22}{12}$

$a = \frac{5}{6}$

$c = - 3$

$b = \frac{- 22}{12}$

$a = \frac{5}{6}$

$c = - 3$

Cancel the common factor of $- 22$ and $12$ .

Factor $2$ out of $- 22$ .

$b = \frac{2 (- 11)}{12}$

$a = \frac{5}{6}$

$c = - 3$

Cancel the common factors.

Factor $2$ out of $12$ .

$b = \frac{2 \cdot - 11}{2 \cdot 6}$

$a = \frac{5}{6}$

$c = - 3$

Cancel the common factor.

$b = \frac{2 \cdot - 11}{2 \cdot 6}$

$a = \frac{5}{6}$

$c = - 3$

Rewrite the expression.

$b = \frac{- 11}{6}$

$a = \frac{5}{6}$

$c = - 3$

$b = \frac{- 11}{6}$

$a = \frac{5}{6}$

$c = - 3$

$b = \frac{- 11}{6}$

$a = \frac{5}{6}$

$c = - 3$

Move the negative in front of the fraction.

$b = - \frac{11}{6}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{11}{6}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{11}{6}$

$a = \frac{5}{6}$

$c = - 3$

$b = - \frac{11}{6}$

$a = \frac{5}{6}$

$c = - 3$

List all of the solutions.

$b = - \frac{11}{6}, a = \frac{5}{6}, c = - 3$

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

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{5}{6}$ , $b = - \frac{11}{6}$ , $c = - 3$ , and $x = 1$ .

Simplify each term.

One to any power is one.

$y = \frac{5}{6} \cdot 1 + (- \frac{11}{6}) \cdot (1) - 3$

Multiply $\frac{5}{6}$ by $1$ .

$y = \frac{5}{6} + (- \frac{11}{6}) \cdot (1) - 3$

Multiply $- 1$ by $1$ .

$y = \frac{5}{6} - \frac{11}{6} - 3$

Combine fractions.

Combine the numerators over the common denominator.

$y = - 3 + \frac{5 - 11}{6}$

Simplify the expression.

Subtract $11$ from $5$ .

$y = - 3 + \frac{- 6}{6}$

Divide $- 6$ by $6$ .

$y = - 3 - 1$

Subtract $1$ from $- 3$ .

$y = - 4$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 1$ . This check passes since $y = - 4$ and $y = - 4$ .

$- 4 = - 4$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{5}{6}$ , $b = - \frac{11}{6}$ , $c = - 3$ , and $x = 3$ .

Simplify each term.

Raise $3$ to the power of $2$ .

$y = \frac{5}{6} \cdot 9 + (- \frac{11}{6}) \cdot (3) - 3$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

$y = \frac{5}{3 (2)} \cdot 9 + (- \frac{11}{6}) \cdot (3) - 3$

Factor $3$ out of $9$ .

$y = \frac{5}{3 \cdot 2} \cdot (3 \cdot 3) + (- \frac{11}{6}) \cdot (3) - 3$

Cancel the common factor.

$y = \frac{5}{3 \cdot 2} \cdot (3 \cdot 3) + (- \frac{11}{6}) \cdot (3) - 3$

Rewrite the expression.

$y = \frac{5}{2} \cdot 3 + (- \frac{11}{6}) \cdot (3) - 3$

Combine $\frac{5}{2}$ and $3$ .

$y = \frac{5 \cdot 3}{2} + (- \frac{11}{6}) \cdot (3) - 3$

Multiply $5$ by $3$ .

$y = \frac{15}{2} + (- \frac{11}{6}) \cdot (3) - 3$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{11}{6}$ into the numerator.

$y = \frac{15}{2} + \frac{- 11}{6} \cdot 3 - 3$

Factor $3$ out of $6$ .

$y = \frac{15}{2} + \frac{- 11}{3 (2)} \cdot 3 - 3$

Cancel the common factor.

$y = \frac{15}{2} + \frac{- 11}{3 \cdot 2} \cdot 3 - 3$

Rewrite the expression.

$y = \frac{15}{2} + \frac{- 11}{2} - 3$

Move the negative in front of the fraction.

$y = \frac{15}{2} - \frac{11}{2} - 3$

Combine fractions.

Combine the numerators over the common denominator.

$y = - 3 + \frac{15 - 11}{2}$

Simplify the expression.

Subtract $11$ from $15$ .

$y = - 3 + \frac{4}{2}$

Divide $4$ by $2$ .

$y = - 3 + 2$

Add $- 3$ and $2$ .

$y = - 1$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 3$ . This check passes since $y = - 1$ and $y = - 1$ .

$- 1 = - 1$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{5}{6}$ , $b = - \frac{11}{6}$ , $c = - 3$ , and $x = 4$ .

Simplify each term.

Raise $4$ to the power of $2$ .

$y = \frac{5}{6} \cdot 16 + (- \frac{11}{6}) \cdot (4) - 3$

Cancel the common factor of $2$ .

Factor $2$ out of $6$ .

$y = \frac{5}{2 (3)} \cdot 16 + (- \frac{11}{6}) \cdot (4) - 3$

Factor $2$ out of $16$ .

$y = \frac{5}{2 \cdot 3} \cdot (2 \cdot 8) + (- \frac{11}{6}) \cdot (4) - 3$

Cancel the common factor.

$y = \frac{5}{2 \cdot 3} \cdot (2 \cdot 8) + (- \frac{11}{6}) \cdot (4) - 3$

Rewrite the expression.

$y = \frac{5}{3} \cdot 8 + (- \frac{11}{6}) \cdot (4) - 3$

Combine $\frac{5}{3}$ and $8$ .

$y = \frac{5 \cdot 8}{3} + (- \frac{11}{6}) \cdot (4) - 3$

Multiply $5$ by $8$ .

$y = \frac{40}{3} + (- \frac{11}{6}) \cdot (4) - 3$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{11}{6}$ into the numerator.

$y = \frac{40}{3} + \frac{- 11}{6} \cdot 4 - 3$

Factor $2$ out of $6$ .

$y = \frac{40}{3} + \frac{- 11}{2 (3)} \cdot 4 - 3$

Factor $2$ out of $4$ .

$y = \frac{40}{3} + \frac{- 11}{2 \cdot 3} \cdot (2 \cdot 2) - 3$

Cancel the common factor.

$y = \frac{40}{3} + \frac{- 11}{2 \cdot 3} \cdot (2 \cdot 2) - 3$

Rewrite the expression.

$y = \frac{40}{3} + \frac{- 11}{3} \cdot 2 - 3$

Combine $\frac{- 11}{3}$ and $2$ .

$y = \frac{40}{3} + \frac{- 11 \cdot 2}{3} - 3$

Multiply $- 11$ by $2$ .

$y = \frac{40}{3} + \frac{- 22}{3} - 3$

Move the negative in front of the fraction.

$y = \frac{40}{3} - \frac{22}{3} - 3$

Combine fractions.

Combine the numerators over the common denominator.

$y = - 3 + \frac{40 - 22}{3}$

Simplify the expression.

Subtract $22$ from $40$ .

$y = - 3 + \frac{18}{3}$

Divide $18$ by $3$ .

$y = - 3 + 6$

Add $- 3$ and $6$ .

$y = 3$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = 3$ and $y = 3$ .

$3 = 3$

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

The function is quadratic

Step 3

Since all $y = y$ , the function is quadratic and follows the form $y = \frac{5 x^{2}}{6} - \frac{11 x}{6} - 3$ .

$y = \frac{5 x^{2}}{6} - \frac{11 x}{6} - 3$

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

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