Algebra Examples

$\begin{array}{c} x & y \\ 2 & \frac{1}{3} \\ 4 & \frac{1}{6} \\ 6 & \frac{1}{9} \end{array}$

Check if the function rule is linear.

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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 $y = a x + b$ .

$\frac{1}{3} = a (2) + b \frac{1}{6} = a (4) + b \frac{1}{9} = a (6) + b$

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

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Solve for $b$ in $\frac{1}{3} = a (2) + b$ .

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Rewrite the equation as $a (2) + b = \frac{1}{3}$ .

$a (2) + b = \frac{1}{3}$

$\frac{1}{6} = a (4) + b$

$\frac{1}{9} = a (6) + b$

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

$2 a + b = \frac{1}{3}$

$\frac{1}{6} = a (4) + b$

$\frac{1}{9} = a (6) + b$

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

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

$\frac{1}{6} = a (4) + b$

$\frac{1}{9} = a (6) + b$

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

$\frac{1}{6} = a (4) + b$

$\frac{1}{9} = a (6) + b$

Replace all occurrences of $b$ with $\frac{1}{3} - 2 a$ in each equation.

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Replace all occurrences of $b$ in $\frac{1}{6} = a (4) + b$ with $\frac{1}{3} - 2 a$ .

$\frac{1}{6} = a (4) + \frac{1}{3} - 2 a$

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

$\frac{1}{9} = a (6) + b$

Simplify $\frac{1}{6} = a (4) + \frac{1}{3} - 2 a$ .

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Simplify the left side.

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Remove parentheses.

$\frac{1}{6} = a (4) + \frac{1}{3} - 2 a$

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

$\frac{1}{9} = a (6) + b$

$\frac{1}{6} = a (4) + \frac{1}{3} - 2 a$

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

$\frac{1}{9} = a (6) + b$

Simplify the right side.

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Simplify $a (4) + \frac{1}{3} - 2 a$ .

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

$\frac{1}{6} = 4 a + \frac{1}{3} - 2 a$

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

$\frac{1}{9} = a (6) + b$

Subtract $2 a$ from $4 a$ .

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = a (6) + b$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = a (6) + b$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = a (6) + b$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = a (6) + b$

Replace all occurrences of $b$ in $\frac{1}{9} = a (6) + b$ with $\frac{1}{3} - 2 a$ .

$\frac{1}{9} = a (6) + \frac{1}{3} - 2 a$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Simplify $\frac{1}{9} = a (6) + \frac{1}{3} - 2 a$ .

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Simplify the left side.

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Remove parentheses.

$\frac{1}{9} = a (6) + \frac{1}{3} - 2 a$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = a (6) + \frac{1}{3} - 2 a$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Simplify the right side.

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Simplify $a (6) + \frac{1}{3} - 2 a$ .

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

$\frac{1}{9} = 6 a + \frac{1}{3} - 2 a$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Subtract $2 a$ from $6 a$ .

$\frac{1}{9} = 4 a + \frac{1}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = 4 a + \frac{1}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = 4 a + \frac{1}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = 4 a + \frac{1}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$\frac{1}{9} = 4 a + \frac{1}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Solve for $a$ in $\frac{1}{9} = 4 a + \frac{1}{3}$ .

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Rewrite the equation as $4 a + \frac{1}{3} = \frac{1}{9}$ .

$4 a + \frac{1}{3} = \frac{1}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

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

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Subtract $\frac{1}{3}$ from both sides of the equation.

$4 a = \frac{1}{9} - \frac{1}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$4 a = \frac{1}{9} - \frac{1}{3} \cdot \frac{3}{3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

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

$4 a = \frac{1}{9} - \frac{3}{3 \cdot 3}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Multiply $3$ by $3$ .

$4 a = \frac{1}{9} - \frac{3}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$4 a = \frac{1}{9} - \frac{3}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Combine the numerators over the common denominator.

$4 a = \frac{1 - 3}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Subtract $3$ from $1$ .

$4 a = \frac{- 2}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Move the negative in front of the fraction.

$4 a = - \frac{2}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$4 a = - \frac{2}{9}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Divide each term in $4 a = - \frac{2}{9}$ by $4$ and simplify.

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Divide each term in $4 a = - \frac{2}{9}$ by $4$ .

$\frac{4 a}{4} = \frac{- \frac{2}{9}}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 a}{4} = \frac{- \frac{2}{9}}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Divide $a$ by $1$ .

$a = \frac{- \frac{2}{9}}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = \frac{- \frac{2}{9}}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = \frac{- \frac{2}{9}}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

$a = - \frac{2}{9} \cdot \frac{1}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{2}{9}$ into the numerator.

$a = \frac{- 2}{9} \cdot \frac{1}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Factor $2$ out of $- 2$ .

$a = \frac{2 (- 1)}{9} \cdot \frac{1}{4}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Factor $2$ out of $4$ .

$a = \frac{2 \cdot - 1}{9} \cdot \frac{1}{2 \cdot 2}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Cancel the common factor.

$a = \frac{2 \cdot - 1}{9} \cdot \frac{1}{2 \cdot 2}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Rewrite the expression.

$a = \frac{- 1}{9} \cdot \frac{1}{2}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = \frac{- 1}{9} \cdot \frac{1}{2}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Multiply $\frac{- 1}{9}$ by $\frac{1}{2}$ .

$a = \frac{- 1}{9 \cdot 2}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Simplify the expression.

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Multiply $9$ by $2$ .

$a = \frac{- 1}{18}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Move the negative in front of the fraction.

$a = - \frac{1}{18}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = - \frac{1}{18}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = - \frac{1}{18}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = - \frac{1}{18}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

$a = - \frac{1}{18}$

$\frac{1}{6} = 2 a + \frac{1}{3}$

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

Replace all occurrences of $a$ with $- \frac{1}{18}$ in each equation.

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Replace all occurrences of $a$ in $\frac{1}{6} = 2 a + \frac{1}{3}$ with $- \frac{1}{18}$ .

$\frac{1}{6} = 2 (- \frac{1}{18}) + \frac{1}{3}$

$a = - \frac{1}{18}$

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

Simplify the right side.

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Simplify $2 (- \frac{1}{18}) + \frac{1}{3}$ .

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Simplify each term.

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Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{18}$ into the numerator.

$\frac{1}{6} = 2 (\frac{- 1}{18}) + \frac{1}{3}$

$a = - \frac{1}{18}$

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

Factor $2$ out of $18$ .

$\frac{1}{6} = 2 (\frac{- 1}{2 (9)}) + \frac{1}{3}$

$a = - \frac{1}{18}$

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

Cancel the common factor.

$\frac{1}{6} = 2 (\frac{- 1}{2 \cdot 9}) + \frac{1}{3}$

$a = - \frac{1}{18}$

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

Rewrite the expression.

$\frac{1}{6} = \frac{- 1}{9} + \frac{1}{3}$

$a = - \frac{1}{18}$

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

$\frac{1}{6} = \frac{- 1}{9} + \frac{1}{3}$

$a = - \frac{1}{18}$

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

Move the negative in front of the fraction.

$\frac{1}{6} = - \frac{1}{9} + \frac{1}{3}$

$a = - \frac{1}{18}$

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

$\frac{1}{6} = - \frac{1}{9} + \frac{1}{3}$

$a = - \frac{1}{18}$

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

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{6} = - \frac{1}{9} + \frac{1}{3} \cdot \frac{3}{3}$

$a = - \frac{1}{18}$

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

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

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

$\frac{1}{6} = - \frac{1}{9} + \frac{3}{3 \cdot 3}$

$a = - \frac{1}{18}$

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

Multiply $3$ by $3$ .

$\frac{1}{6} = - \frac{1}{9} + \frac{3}{9}$

$a = - \frac{1}{18}$

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

$\frac{1}{6} = - \frac{1}{9} + \frac{3}{9}$

$a = - \frac{1}{18}$

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

Combine the numerators over the common denominator.

$\frac{1}{6} = \frac{- 1 + 3}{9}$

$a = - \frac{1}{18}$

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

Add $- 1$ and $3$ .

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

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

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

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

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

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

Replace all occurrences of $a$ in $b = \frac{1}{3} - 2 a$ with $- \frac{1}{18}$ .

$b = \frac{1}{3} - 2 (- \frac{1}{18})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Simplify the right side.

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Simplify $\frac{1}{3} - 2 (- \frac{1}{18})$ .

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Simplify each term.

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Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{18}$ into the numerator.

$b = \frac{1}{3} - 2 (\frac{- 1}{18})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Factor $2$ out of $- 2$ .

$b = \frac{1}{3} + 2 (- 1) (\frac{- 1}{18})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Factor $2$ out of $18$ .

$b = \frac{1}{3} + 2 \cdot (- 1 \frac{- 1}{2 \cdot 9})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Cancel the common factor.

$b = \frac{1}{3} + 2 \cdot (- 1 \frac{- 1}{2 \cdot 9})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Rewrite the expression.

$b = \frac{1}{3} - 1 (\frac{- 1}{9})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{1}{3} - 1 (\frac{- 1}{9})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Move the negative in front of the fraction.

$b = \frac{1}{3} - 1 (- \frac{1}{9})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Multiply $- 1 (- \frac{1}{9})$ .

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Multiply $- 1$ by $- 1$ .

$b = \frac{1}{3} + 1 (\frac{1}{9})$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Multiply $\frac{1}{9}$ by $1$ .

$b = \frac{1}{3} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{1}{3} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{1}{3} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$b = \frac{1}{3} \cdot \frac{3}{3} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

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

$b = \frac{3}{3 \cdot 3} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Multiply $3$ by $3$ .

$b = \frac{3}{9} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{3}{9} + \frac{1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Combine the numerators over the common denominator.

$b = \frac{3 + 1}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Add $3$ and $1$ .

$b = \frac{4}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{4}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{4}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

$b = \frac{4}{9}$

$\frac{1}{6} = \frac{2}{9}$

$a = - \frac{1}{18}$

Since $\frac{1}{6} = \frac{2}{9}$ 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

Check if the function rule is quadratic.

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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 $c$ .

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Solve for $c$ in $\frac{1}{3} = a \cdot 2^{2} + b (2) + c$ .

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Rewrite the equation as $a \cdot 2^{2} + b (2) + c = \frac{1}{3}$ .

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

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Simplify each term.

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

$a \cdot 4 + b (2) + c = \frac{1}{3}$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

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

$4 \cdot a + b (2) + c = \frac{1}{3}$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

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

$4 a + 2 b + c = \frac{1}{3}$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$4 a + 2 b + c = \frac{1}{3}$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

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

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

$2 b + c = \frac{1}{3} - 4 a$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

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

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + c$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Replace all occurrences of $c$ with $\frac{1}{3} - 4 a - 2 b$ in each equation.

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Replace all occurrences of $c$ in $\frac{1}{6} = a \cdot 4^{2} + b (4) + c$ with $\frac{1}{3} - 4 a - 2 b$ .

$\frac{1}{6} = a \cdot 4^{2} + b (4) + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Simplify $\frac{1}{6} = a \cdot 4^{2} + b (4) + \frac{1}{3} - 4 a - 2 b$ .

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Simplify the left side.

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Remove parentheses.

$\frac{1}{6} = a \cdot 4^{2} + b (4) + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$\frac{1}{6} = a \cdot 4^{2} + b (4) + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Simplify the right side.

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Simplify $a \cdot 4^{2} + b (4) + \frac{1}{3} - 4 a - 2 b$ .

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Simplify each term.

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

$\frac{1}{6} = a \cdot 16 + b (4) + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Move $16$ to the left of $a$ .

$\frac{1}{6} = 16 \cdot a + b (4) + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

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

$\frac{1}{6} = 16 a + 4 b + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$\frac{1}{6} = 16 a + 4 b + \frac{1}{3} - 4 a - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Simplify by adding terms.

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Subtract $4 a$ from $16 a$ .

$\frac{1}{6} = 12 a + 4 b + \frac{1}{3} - 2 b$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Subtract $2 b$ from $4 b$ .

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + c$

Replace all occurrences of $c$ in $\frac{1}{9} = a \cdot 6^{2} + b (6) + c$ with $\frac{1}{3} - 4 a - 2 b$ .

$\frac{1}{9} = a \cdot 6^{2} + b (6) + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify $\frac{1}{9} = a \cdot 6^{2} + b (6) + \frac{1}{3} - 4 a - 2 b$ .

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Simplify the left side.

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Remove parentheses.

$\frac{1}{9} = a \cdot 6^{2} + b (6) + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = a \cdot 6^{2} + b (6) + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify the right side.

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Simplify $a \cdot 6^{2} + b (6) + \frac{1}{3} - 4 a - 2 b$ .

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Simplify each term.

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Raise $6$ to the power of $2$ .

$\frac{1}{9} = a \cdot 36 + b (6) + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Move $36$ to the left of $a$ .

$\frac{1}{9} = 36 \cdot a + b (6) + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

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

$\frac{1}{9} = 36 a + 6 b + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = 36 a + 6 b + \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify by adding terms.

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Subtract $4 a$ from $36 a$ .

$\frac{1}{9} = 32 a + 6 b + \frac{1}{3} - 2 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Subtract $2 b$ from $6 b$ .

$\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Solve for $a$ in $\frac{1}{9} = 32 a + 4 b + \frac{1}{3}$ .

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Rewrite the equation as $32 a + 4 b + \frac{1}{3} = \frac{1}{9}$ .

$32 a + 4 b + \frac{1}{3} = \frac{1}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

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

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

$32 a + \frac{1}{3} = \frac{1}{9} - 4 b$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Subtract $\frac{1}{3}$ from both sides of the equation.

$32 a = \frac{1}{9} - 4 b - \frac{1}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$32 a = - 4 b + \frac{1}{9} - \frac{1}{3} \cdot \frac{3}{3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

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

$32 a = - 4 b + \frac{1}{9} - \frac{3}{3 \cdot 3}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $3$ by $3$ .

$32 a = - 4 b + \frac{1}{9} - \frac{3}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$32 a = - 4 b + \frac{1}{9} - \frac{3}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Combine the numerators over the common denominator.

$32 a = - 4 b + \frac{1 - 3}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Subtract $3$ from $1$ .

$32 a = - 4 b + \frac{- 2}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Move the negative in front of the fraction.

$32 a = - 4 b - \frac{2}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$32 a = - 4 b - \frac{2}{9}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Divide each term in $32 a = - 4 b - \frac{2}{9}$ by $32$ and simplify.

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Divide each term in $32 a = - 4 b - \frac{2}{9}$ by $32$ .

$\frac{32 a}{32} = \frac{- 4 b}{32} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify the left side.

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Cancel the common factor of $32$ .

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Cancel the common factor.

$\frac{32 a}{32} = \frac{- 4 b}{32} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Divide $a$ by $1$ .

$a = \frac{- 4 b}{32} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = \frac{- 4 b}{32} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = \frac{- 4 b}{32} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify the right side.

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Simplify each term.

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Cancel the common factor of $- 4$ and $32$ .

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Factor $4$ out of $- 4 b$ .

$a = \frac{4 (- b)}{32} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factors.

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Factor $4$ out of $32$ .

$a = \frac{4 (- b)}{4 (8)} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor.

$a = \frac{4 (- b)}{4 \cdot 8} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Rewrite the expression.

$a = \frac{- b}{8} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = \frac{- b}{8} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = \frac{- b}{8} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Move the negative in front of the fraction.

$a = - \frac{b}{8} + \frac{- \frac{2}{9}}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply the numerator by the reciprocal of the denominator.

$a = - \frac{b}{8} - \frac{2}{9} \cdot \frac{1}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{2}{9}$ into the numerator.

$a = - \frac{b}{8} + \frac{- 2}{9} \cdot \frac{1}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $2$ out of $- 2$ .

$a = - \frac{b}{8} + \frac{2 (- 1)}{9} \cdot \frac{1}{32}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $2$ out of $32$ .

$a = - \frac{b}{8} + \frac{2 \cdot - 1}{9} \cdot \frac{1}{2 \cdot 16}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor.

$a = - \frac{b}{8} + \frac{2 \cdot - 1}{9} \cdot \frac{1}{2 \cdot 16}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Rewrite the expression.

$a = - \frac{b}{8} + \frac{- 1}{9} \cdot \frac{1}{16}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = - \frac{b}{8} + \frac{- 1}{9} \cdot \frac{1}{16}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $\frac{- 1}{9}$ by $\frac{1}{16}$ .

$a = - \frac{b}{8} + \frac{- 1}{9 \cdot 16}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $9$ by $16$ .

$a = - \frac{b}{8} + \frac{- 1}{144}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Move the negative in front of the fraction.

$a = - \frac{b}{8} - \frac{1}{144}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = - \frac{b}{8} - \frac{1}{144}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = - \frac{b}{8} - \frac{1}{144}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = - \frac{b}{8} - \frac{1}{144}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

$a = - \frac{b}{8} - \frac{1}{144}$

$\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$

$c = \frac{1}{3} - 4 a - 2 b$

Replace all occurrences of $a$ with $- \frac{b}{8} - \frac{1}{144}$ in each equation.

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Replace all occurrences of $a$ in $\frac{1}{6} = 12 a + 2 b + \frac{1}{3}$ with $- \frac{b}{8} - \frac{1}{144}$ .

$\frac{1}{6} = 12 (- \frac{b}{8} - \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify the right side.

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Simplify $12 (- \frac{b}{8} - \frac{1}{144}) + 2 b + \frac{1}{3}$ .

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Simplify each term.

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Apply the distributive property.

$\frac{1}{6} = 12 (- \frac{b}{8}) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor of $4$ .

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Move the leading negative in $- \frac{b}{8}$ into the numerator.

$\frac{1}{6} = 12 (\frac{- b}{8}) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $4$ out of $12$ .

$\frac{1}{6} = 4 (3) (\frac{- b}{8}) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $4$ out of $8$ .

$\frac{1}{6} = 4 \cdot (3 (\frac{- b}{4 \cdot 2})) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor.

$\frac{1}{6} = 4 \cdot (3 (\frac{- b}{4 \cdot 2})) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Rewrite the expression.

$\frac{1}{6} = 3 (\frac{- b}{2}) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = 3 (\frac{- b}{2}) + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

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

$\frac{1}{6} = \frac{3 (- b)}{2} + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $- 1$ by $3$ .

$\frac{1}{6} = \frac{- 3 b}{2} + 12 (- \frac{1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor of $12$ .

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Move the leading negative in $- \frac{1}{144}$ into the numerator.

$\frac{1}{6} = \frac{- 3 b}{2} + 12 (\frac{- 1}{144}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $12$ out of $144$ .

$\frac{1}{6} = \frac{- 3 b}{2} + 12 (\frac{- 1}{12 (12)}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor.

$\frac{1}{6} = \frac{- 3 b}{2} + 12 (\frac{- 1}{12 \cdot 12}) + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Rewrite the expression.

$\frac{1}{6} = \frac{- 3 b}{2} + \frac{- 1}{12} + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{- 3 b}{2} + \frac{- 1}{12} + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify each term.

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Move the negative in front of the fraction.

$\frac{1}{6} = - \frac{(3) b}{2} + \frac{- 1}{12} + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Move the negative in front of the fraction.

$\frac{1}{6} = - \frac{3 b}{2} - \frac{1}{12} + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = - \frac{3 b}{2} - \frac{1}{12} + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = - \frac{3 b}{2} - \frac{1}{12} + 2 b + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

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

$\frac{1}{6} = - \frac{3 b}{2} + 2 b \cdot \frac{2}{2} - \frac{1}{12} + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Combine $2 b$ and $\frac{2}{2}$ .

$\frac{1}{6} = - \frac{3 b}{2} + \frac{2 b \cdot 2}{2} - \frac{1}{12} + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Combine the numerators over the common denominator.

$\frac{1}{6} = \frac{- 3 b + 2 b \cdot 2}{2} - \frac{1}{12} + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Find the common denominator.

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Multiply $\frac{- 3 b + 2 b \cdot 2}{2}$ by $\frac{6}{6}$ .

$\frac{1}{6} = \frac{- 3 b + 2 b \cdot 2}{2} \cdot \frac{6}{6} - \frac{1}{12} + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $\frac{- 3 b + 2 b \cdot 2}{2}$ by $\frac{6}{6}$ .

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6}{2 \cdot 6} - \frac{1}{12} + \frac{1}{3}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $\frac{1}{3}$ by $\frac{4}{4}$ .

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6}{2 \cdot 6} - \frac{1}{12} + \frac{1}{3} \cdot \frac{4}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $\frac{1}{3}$ by $\frac{4}{4}$ .

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6}{2 \cdot 6} - \frac{1}{12} + \frac{4}{3 \cdot 4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $2$ by $6$ .

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6}{12} - \frac{1}{12} + \frac{4}{3 \cdot 4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Multiply $3$ by $4$ .

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6}{12} - \frac{1}{12} + \frac{4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6}{12} - \frac{1}{12} + \frac{4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Combine the numerators over the common denominator.

$\frac{1}{6} = \frac{(- 3 b + 2 b \cdot 2) \cdot 6 - 1 + 4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Simplify each term.

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Multiply $2$ by $2$ .

$\frac{1}{6} = \frac{(- 3 b + 4 b) \cdot 6 - 1 + 4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Add $- 3 b$ and $4 b$ .

$\frac{1}{6} = \frac{b \cdot 6 - 1 + 4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

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

$\frac{1}{6} = \frac{6 b - 1 + 4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{6 b - 1 + 4}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Reduce the expression by cancelling the common factors.

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Add $- 1$ and $4$ .

$\frac{1}{6} = \frac{6 b + 3}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor of $6 b + 3$ and $12$ .

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Factor $3$ out of $6 b$ .

$\frac{1}{6} = \frac{3 (2 b) + 3}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $3$ out of $3$ .

$\frac{1}{6} = \frac{3 (2 b) + 3 (1)}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Factor $3$ out of $3 (2 b) + 3 (1)$ .

$\frac{1}{6} = \frac{3 (2 b + 1)}{12}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factors.

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Factor $3$ out of $12$ .

$\frac{1}{6} = \frac{3 (2 b + 1)}{3 (4)}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Cancel the common factor.

$\frac{1}{6} = \frac{3 (2 b + 1)}{3 \cdot 4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Rewrite the expression.

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 4 a - 2 b$

Replace all occurrences of $a$ in $c = \frac{1}{3} - 4 a - 2 b$ with $- \frac{b}{8} - \frac{1}{144}$ .

$c = \frac{1}{3} - 4 (- \frac{b}{8} - \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the right side.

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Simplify $\frac{1}{3} - 4 (- \frac{b}{8} - \frac{1}{144}) - 2 b$ .

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Simplify each term.

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Apply the distributive property.

$c = \frac{1}{3} - 4 (- \frac{b}{8}) - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor of $4$ .

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Move the leading negative in $- \frac{b}{8}$ into the numerator.

$c = \frac{1}{3} - 4 \frac{- b}{8} - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $4$ out of $- 4$ .

$c = \frac{1}{3} + 4 (- 1) (\frac{- b}{8}) - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $4$ out of $8$ .

$c = \frac{1}{3} + 4 \cdot (- 1 \frac{- b}{4 \cdot 2}) - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor.

$c = \frac{1}{3} + 4 \cdot (- 1 \frac{- b}{4 \cdot 2}) - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Rewrite the expression.

$c = \frac{1}{3} - 1 \frac{- b}{2} - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 1 \frac{- b}{2} - 4 (- \frac{1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor of $4$ .

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Move the leading negative in $- \frac{1}{144}$ into the numerator.

$c = \frac{1}{3} - 1 \frac{- b}{2} - 4 (\frac{- 1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $4$ out of $- 4$ .

$c = \frac{1}{3} - 1 \frac{- b}{2} + 4 (- 1) (\frac{- 1}{144}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $4$ out of $144$ .

$c = \frac{1}{3} - 1 \frac{- b}{2} + 4 \cdot (- 1 \frac{- 1}{4 \cdot 36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor.

$c = \frac{1}{3} - 1 \frac{- b}{2} + 4 \cdot (- 1 \frac{- 1}{4 \cdot 36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Rewrite the expression.

$c = \frac{1}{3} - 1 \frac{- b}{2} - 1 (\frac{- 1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} - 1 \frac{- b}{2} - 1 (\frac{- 1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify each term.

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Move the negative in front of the fraction.

$c = \frac{1}{3} - 1 (- \frac{b}{2}) - 1 (\frac{- 1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

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Multiply $- 1$ by $- 1$ .

$c = \frac{1}{3} + 1 (\frac{b}{2}) - 1 (\frac{- 1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = \frac{1}{3} + \frac{b}{2} - 1 (\frac{- 1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} + \frac{b}{2} - 1 (\frac{- 1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Move the negative in front of the fraction.

$c = \frac{1}{3} + \frac{b}{2} - 1 (- \frac{1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $- 1 (- \frac{1}{36})$ .

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Multiply $- 1$ by $- 1$ .

$c = \frac{1}{3} + \frac{b}{2} + 1 (\frac{1}{36}) - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $\frac{1}{36}$ by $1$ .

$c = \frac{1}{3} + \frac{b}{2} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} + \frac{b}{2} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} + \frac{b}{2} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{3} + \frac{b}{2} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$c = \frac{b}{2} + \frac{1}{3} \cdot \frac{12}{12} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{3}$ by $\frac{12}{12}$ .

$c = \frac{b}{2} + \frac{12}{3 \cdot 12} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $3$ by $12$ .

$c = \frac{b}{2} + \frac{12}{36} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{b}{2} + \frac{12}{36} + \frac{1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Combine the numerators over the common denominator.

$c = \frac{b}{2} + \frac{12 + 1}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Add $12$ and $1$ .

$c = \frac{b}{2} + \frac{13}{36} - 2 b$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = \frac{b}{2} - 2 b \cdot \frac{2}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify terms.

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

$c = \frac{b}{2} + \frac{- 2 b \cdot 2}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Combine the numerators over the common denominator.

$c = \frac{b - 2 b \cdot 2}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{b - 2 b \cdot 2}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify each term.

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Simplify the numerator.

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Factor $b$ out of $b - 2 b \cdot 2$ .

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Raise $b$ to the power of $1$ .

$c = \frac{b - 2 b \cdot 2}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $b$ out of $b^{1}$ .

$c = \frac{b \cdot 1 - 2 b \cdot 2}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $b$ out of $- 2 b \cdot 2$ .

$c = \frac{b \cdot 1 + b (- 2 \cdot 2)}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $b$ out of $b \cdot 1 + b (- 2 \cdot 2)$ .

$c = \frac{b (1 - 2 \cdot 2)}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{b (1 - 2 \cdot 2)}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $- 2$ by $2$ .

$c = \frac{b (1 - 4)}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Subtract $4$ from $1$ .

$c = \frac{b \cdot - 3}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{b \cdot - 3}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = \frac{- 3 \cdot b}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Move the negative in front of the fraction.

$c = - \frac{3 b}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$\frac{1}{6} = \frac{2 b + 1}{4}$

$a = - \frac{b}{8} - \frac{1}{144}$

Solve for $b$ in $\frac{1}{6} = \frac{2 b + 1}{4}$ .

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Rewrite the equation as $\frac{2 b + 1}{4} = \frac{1}{6}$ .

$\frac{2 b + 1}{4} = \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply both sides by $4$ .

$\frac{2 b + 1}{4} \cdot 4 = \frac{1}{6} \cdot 4$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify.

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Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{2 b + 1}{4} \cdot 4 = \frac{1}{6} \cdot 4$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Rewrite the expression.

$2 b + 1 = \frac{1}{6} \cdot 4$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$2 b + 1 = \frac{1}{6} \cdot 4$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$2 b + 1 = \frac{1}{6} \cdot 4$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the right side.

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Simplify $\frac{1}{6} \cdot 4$ .

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Cancel the common factor of $2$ .

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Factor $2$ out of $6$ .

$2 b + 1 = \frac{1}{2 (3)} \cdot 4$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Factor $2$ out of $4$ .

$2 b + 1 = \frac{1}{2 \cdot 3} \cdot (2 \cdot 2)$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor.

$2 b + 1 = \frac{1}{2 \cdot 3} \cdot (2 \cdot 2)$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Rewrite the expression.

$2 b + 1 = \frac{1}{3} \cdot 2$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$2 b + 1 = \frac{1}{3} \cdot 2$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Combine $\frac{1}{3}$ and $2$ .

$2 b + 1 = \frac{2}{3}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$2 b + 1 = \frac{2}{3}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$2 b + 1 = \frac{2}{3}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$2 b + 1 = \frac{2}{3}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Solve for $b$ .

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

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Subtract $1$ from both sides of the equation.

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Combine $- 1$ and $\frac{3}{3}$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Combine the numerators over the common denominator.

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the numerator.

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Multiply $- 1$ by $3$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Subtract $3$ from $2$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Move the negative in front of the fraction.

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Divide each term in $2 b = - \frac{1}{3}$ by $2$ and simplify.

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Divide each term in $2 b = - \frac{1}{3}$ by $2$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Divide $b$ by $1$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the right side.

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Multiply the numerator by the reciprocal of the denominator.

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $- \frac{1}{3} \cdot \frac{1}{2}$ .

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Multiply $\frac{1}{2}$ by $\frac{1}{3}$ .

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

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $2$ by $3$ .

$b = - \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$b = - \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$b = - \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$b = - \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$b = - \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

$b = - \frac{1}{6}$

$c = - \frac{3 b}{2} + \frac{13}{36}$

$a = - \frac{b}{8} - \frac{1}{144}$

Replace all occurrences of $b$ with $- \frac{1}{6}$ in each equation.

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Replace all occurrences of $b$ in $c = - \frac{3 b}{2} + \frac{13}{36}$ with $- \frac{1}{6}$ .

$c = - \frac{3 (- \frac{1}{6})}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the right side.

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Simplify $- \frac{3 (- \frac{1}{6})}{2} + \frac{13}{36}$ .

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Simplify each term.

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Simplify the numerator.

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Multiply $- 1$ by $3$ .

$c = - \frac{- 3 (\frac{1}{6})}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Combine $- 3$ and $\frac{1}{6}$ .

$c = - \frac{\frac{- 3}{6}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{\frac{- 3}{6}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor of $- 3$ and $6$ .

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Factor $3$ out of $- 3$ .

$c = - \frac{\frac{3 (- 1)}{6}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factors.

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Factor $3$ out of $6$ .

$c = - \frac{\frac{3 \cdot - 1}{3 \cdot 2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor.

$c = - \frac{\frac{3 \cdot - 1}{3 \cdot 2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Rewrite the expression.

$c = - \frac{\frac{- 1}{2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{\frac{- 1}{2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{\frac{- 1}{2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Move the negative in front of the fraction.

$c = - \frac{- \frac{1}{2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = - \frac{- \frac{1}{2}}{2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply the numerator by the reciprocal of the denominator.

$c = - (- \frac{1}{2} \cdot \frac{1}{2}) + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $- \frac{1}{2} \cdot \frac{1}{2}$ .

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

$c = \frac{1}{2 \cdot 2} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $2$ by $2$ .

$c = \frac{1}{4} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{4} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $- - \frac{1}{4}$ .

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Multiply $- 1$ by $- 1$ .

$c = 1 (\frac{1}{4}) + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

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

$c = \frac{1}{4} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{4} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{1}{4} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$c = \frac{1}{4} \cdot \frac{9}{9} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{4}$ by $\frac{9}{9}$ .

$c = \frac{9}{4 \cdot 9} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Multiply $4$ by $9$ .

$c = \frac{9}{36} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{9}{36} + \frac{13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Simplify the expression.

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Combine the numerators over the common denominator.

$c = \frac{9 + 13}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Add $9$ and $13$ .

$c = \frac{22}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{22}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor of $22$ and $36$ .

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Factor $2$ out of $22$ .

$c = \frac{2 (11)}{36}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factors.

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Factor $2$ out of $36$ .

$c = \frac{2 \cdot 11}{2 \cdot 18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Cancel the common factor.

$c = \frac{2 \cdot 11}{2 \cdot 18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Rewrite the expression.

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = - \frac{b}{8} - \frac{1}{144}$

Replace all occurrences of $b$ in $a = - \frac{b}{8} - \frac{1}{144}$ with $- \frac{1}{6}$ .

$a = - \frac{- \frac{1}{6}}{8} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Simplify the right side.

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Simplify $- \frac{- \frac{1}{6}}{8} - \frac{1}{144}$ .

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Simplify each term.

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Multiply the numerator by the reciprocal of the denominator.

$a = - (- \frac{1}{6} \cdot \frac{1}{8}) - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Multiply $- \frac{1}{6} \cdot \frac{1}{8}$ .

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Multiply $\frac{1}{8}$ by $\frac{1}{6}$ .

$a = \frac{1}{8 \cdot 6} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Multiply $8$ by $6$ .

$a = \frac{1}{48} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{48} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Multiply $- - \frac{1}{48}$ .

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Multiply $- 1$ by $- 1$ .

$a = 1 (\frac{1}{48}) - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Multiply $\frac{1}{48}$ by $1$ .

$a = \frac{1}{48} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{48} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{48} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

To write $\frac{1}{48}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = \frac{1}{48} \cdot \frac{3}{3} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Write each expression with a common denominator of $144$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{48}$ by $\frac{3}{3}$ .

$a = \frac{3}{48 \cdot 3} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Multiply $48$ by $3$ .

$a = \frac{3}{144} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{3}{144} - \frac{1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Simplify the expression.

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Combine the numerators over the common denominator.

$a = \frac{3 - 1}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Subtract $1$ from $3$ .

$a = \frac{2}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{2}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Cancel the common factor of $2$ and $144$ .

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Factor $2$ out of $2$ .

$a = \frac{2 (1)}{144}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Cancel the common factors.

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Factor $2$ out of $144$ .

$a = \frac{2 \cdot 1}{2 \cdot 72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Cancel the common factor.

$a = \frac{2 \cdot 1}{2 \cdot 72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

Rewrite the expression.

$a = \frac{1}{72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

$a = \frac{1}{72}$

$c = \frac{11}{18}$

$b = - \frac{1}{6}$

List all of the solutions.

$a = \frac{1}{72}, c = \frac{11}{18}, b = - \frac{1}{6}$

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

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Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{1}{72}$ , $b = - \frac{1}{6}$ , $c = \frac{11}{18}$ , and $x = 2$ .

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Simplify each term.

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

$y = \frac{1}{72} \cdot 4 + (- \frac{1}{6}) \cdot (2) + \frac{11}{18}$

Cancel the common factor of $4$ .

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Factor $4$ out of $72$ .

$y = \frac{1}{4 (18)} \cdot 4 + (- \frac{1}{6}) \cdot (2) + \frac{11}{18}$

Cancel the common factor.

$y = \frac{1}{4 \cdot 18} \cdot 4 + (- \frac{1}{6}) \cdot (2) + \frac{11}{18}$

Rewrite the expression.

$y = \frac{1}{18} + (- \frac{1}{6}) \cdot (2) + \frac{11}{18}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{6}$ into the numerator.

$y = \frac{1}{18} + \frac{- 1}{6} \cdot 2 + \frac{11}{18}$

Factor $2$ out of $6$ .

$y = \frac{1}{18} + \frac{- 1}{2 (3)} \cdot 2 + \frac{11}{18}$

Cancel the common factor.

$y = \frac{1}{18} + \frac{- 1}{2 \cdot 3} \cdot 2 + \frac{11}{18}$

Rewrite the expression.

$y = \frac{1}{18} + \frac{- 1}{3} + \frac{11}{18}$

Move the negative in front of the fraction.

$y = \frac{1}{18} - \frac{1}{3} + \frac{11}{18}$

Combine fractions.

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Combine the numerators over the common denominator.

$y = \frac{1 + 11}{18} + \frac{- 1}{3}$

Add $1$ and $11$ .

$y = \frac{12}{18} + \frac{- 1}{3}$

Simplify each term.

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Cancel the common factor of $12$ and $18$ .

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Factor $6$ out of $12$ .

$y = \frac{6 (2)}{18} + \frac{- 1}{3}$

Cancel the common factors.

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Factor $6$ out of $18$ .

$y = \frac{6 \cdot 2}{6 \cdot 3} + \frac{- 1}{3}$

Cancel the common factor.

$y = \frac{6 \cdot 2}{6 \cdot 3} + \frac{- 1}{3}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Combine fractions.

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Combine the numerators over the common denominator.

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

Subtract $1$ from $2$ .

$y = \frac{1}{3}$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 2$ . This check passes since $y = \frac{1}{3}$ and $y = \frac{1}{3}$ .

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

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{1}{72}$ , $b = - \frac{1}{6}$ , $c = \frac{11}{18}$ , and $x = 4$ .

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Simplify each term.

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

$y = \frac{1}{72} \cdot 16 + (- \frac{1}{6}) \cdot (4) + \frac{11}{18}$

Cancel the common factor of $8$ .

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Factor $8$ out of $72$ .

$y = \frac{1}{8 (9)} \cdot 16 + (- \frac{1}{6}) \cdot (4) + \frac{11}{18}$

Factor $8$ out of $16$ .

$y = \frac{1}{8 \cdot 9} \cdot (8 \cdot 2) + (- \frac{1}{6}) \cdot (4) + \frac{11}{18}$

Cancel the common factor.

$y = \frac{1}{8 \cdot 9} \cdot (8 \cdot 2) + (- \frac{1}{6}) \cdot (4) + \frac{11}{18}$

Rewrite the expression.

$y = \frac{1}{9} \cdot 2 + (- \frac{1}{6}) \cdot (4) + \frac{11}{18}$

Combine $\frac{1}{9}$ and $2$ .

$y = \frac{2}{9} + (- \frac{1}{6}) \cdot (4) + \frac{11}{18}$

Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{6}$ into the numerator.

$y = \frac{2}{9} + \frac{- 1}{6} \cdot 4 + \frac{11}{18}$

Factor $2$ out of $6$ .

$y = \frac{2}{9} + \frac{- 1}{2 (3)} \cdot 4 + \frac{11}{18}$

Factor $2$ out of $4$ .

$y = \frac{2}{9} + \frac{- 1}{2 \cdot 3} \cdot (2 \cdot 2) + \frac{11}{18}$

Cancel the common factor.

$y = \frac{2}{9} + \frac{- 1}{2 \cdot 3} \cdot (2 \cdot 2) + \frac{11}{18}$

Rewrite the expression.

$y = \frac{2}{9} + \frac{- 1}{3} \cdot 2 + \frac{11}{18}$

Combine $\frac{- 1}{3}$ and $2$ .

$y = \frac{2}{9} + \frac{- 1 \cdot 2}{3} + \frac{11}{18}$

Multiply $- 1$ by $2$ .

$y = \frac{2}{9} + \frac{- 2}{3} + \frac{11}{18}$

Move the negative in front of the fraction.

$y = \frac{2}{9} - \frac{2}{3} + \frac{11}{18}$

Find the common denominator.

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Multiply $\frac{2}{9}$ by $\frac{2}{2}$ .

$y = \frac{2}{9} \cdot \frac{2}{2} - \frac{2}{3} + \frac{11}{18}$

Multiply $\frac{2}{9}$ by $\frac{2}{2}$ .

$y = \frac{2 \cdot 2}{9 \cdot 2} - \frac{2}{3} + \frac{11}{18}$

Multiply $\frac{2}{3}$ by $\frac{6}{6}$ .

$y = \frac{2 \cdot 2}{9 \cdot 2} - (\frac{2}{3} \cdot \frac{6}{6}) + \frac{11}{18}$

Multiply $\frac{2}{3}$ by $\frac{6}{6}$ .

$y = \frac{2 \cdot 2}{9 \cdot 2} - \frac{2 \cdot 6}{3 \cdot 6} + \frac{11}{18}$

Reorder the factors of $9 \cdot 2$ .

$y = \frac{2 \cdot 2}{2 \cdot 9} - \frac{2 \cdot 6}{3 \cdot 6} + \frac{11}{18}$

Multiply $2$ by $9$ .

$y = \frac{2 \cdot 2}{18} - \frac{2 \cdot 6}{3 \cdot 6} + \frac{11}{18}$

Multiply $3$ by $6$ .

$y = \frac{2 \cdot 2}{18} - \frac{2 \cdot 6}{18} + \frac{11}{18}$

Combine the numerators over the common denominator.

$y = \frac{2 \cdot 2 - 2 \cdot 6 + 11}{18}$

Simplify each term.

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Multiply $2$ by $2$ .

$y = \frac{4 - 2 \cdot 6 + 11}{18}$

Multiply $- 2$ by $6$ .

$y = \frac{4 - 12 + 11}{18}$

Reduce the expression by cancelling the common factors.

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Subtract $12$ from $4$ .

$y = \frac{- 8 + 11}{18}$

Add $- 8$ and $11$ .

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

Cancel the common factor of $3$ and $18$ .

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Factor $3$ out of $3$ .

$y = \frac{3 (1)}{18}$

Cancel the common factors.

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Factor $3$ out of $18$ .

$y = \frac{3 \cdot 1}{3 \cdot 6}$

Cancel the common factor.

$y = \frac{3 \cdot 1}{3 \cdot 6}$

Rewrite the expression.

$y = \frac{1}{6}$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 4$ . This check passes since $y = \frac{1}{6}$ and $y = \frac{1}{6}$ .

$\frac{1}{6} = \frac{1}{6}$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = \frac{1}{72}$ , $b = - \frac{1}{6}$ , $c = \frac{11}{18}$ , and $x = 6$ .

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Simplify each term.

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Raise $6$ to the power of $2$ .

$y = \frac{1}{72} \cdot 36 + (- \frac{1}{6}) \cdot (6) + \frac{11}{18}$

Cancel the common factor of $36$ .

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Factor $36$ out of $72$ .

$y = \frac{1}{36 (2)} \cdot 36 + (- \frac{1}{6}) \cdot (6) + \frac{11}{18}$

Cancel the common factor.

$y = \frac{1}{36 \cdot 2} \cdot 36 + (- \frac{1}{6}) \cdot (6) + \frac{11}{18}$

Rewrite the expression.

$y = \frac{1}{2} + (- \frac{1}{6}) \cdot (6) + \frac{11}{18}$

Cancel the common factor of $6$ .

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Move the leading negative in $- \frac{1}{6}$ into the numerator.

$y = \frac{1}{2} + \frac{- 1}{6} \cdot 6 + \frac{11}{18}$

Cancel the common factor.

$y = \frac{1}{2} + \frac{- 1}{6} \cdot 6 + \frac{11}{18}$

Rewrite the expression.

$y = \frac{1}{2} - 1 + \frac{11}{18}$

Find the common denominator.

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Multiply $\frac{1}{2}$ by $\frac{9}{9}$ .

$y = \frac{1}{2} \cdot \frac{9}{9} - 1 + \frac{11}{18}$

Multiply $\frac{1}{2}$ by $\frac{9}{9}$ .

$y = \frac{9}{2 \cdot 9} - 1 + \frac{11}{18}$

Write $- 1$ as a fraction with denominator $1$ .

$y = \frac{9}{2 \cdot 9} + \frac{- 1}{1} + \frac{11}{18}$

Multiply $\frac{- 1}{1}$ by $\frac{18}{18}$ .

$y = \frac{9}{2 \cdot 9} + \frac{- 1}{1} \cdot \frac{18}{18} + \frac{11}{18}$

Multiply $\frac{- 1}{1}$ by $\frac{18}{18}$ .

$y = \frac{9}{2 \cdot 9} + \frac{- 1 \cdot 18}{18} + \frac{11}{18}$

Multiply $2$ by $9$ .

$y = \frac{9}{18} + \frac{- 1 \cdot 18}{18} + \frac{11}{18}$

Combine the numerators over the common denominator.

$y = \frac{9 - 1 \cdot 18 + 11}{18}$

Simplify the expression.

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Multiply $- 1$ by $18$ .

$y = \frac{9 - 18 + 11}{18}$

Subtract $18$ from $9$ .

$y = \frac{- 9 + 11}{18}$

Add $- 9$ and $11$ .

$y = \frac{2}{18}$

Cancel the common factor of $2$ and $18$ .

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Factor $2$ out of $2$ .

$y = \frac{2 (1)}{18}$

Cancel the common factors.

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Factor $2$ out of $18$ .

$y = \frac{2 \cdot 1}{2 \cdot 9}$

Cancel the common factor.

$y = \frac{2 \cdot 1}{2 \cdot 9}$

Rewrite the expression.

$y = \frac{1}{9}$

If the table has a quadratic function rule, $y = y$ for the corresponding $x$ value, $x = 6$ . This check passes since $y = \frac{1}{9}$ and $y = \frac{1}{9}$ .

$\frac{1}{9} = \frac{1}{9}$

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

The function is quadratic

Since all $y = y$ , the function is quadratic and follows the form $y = \frac{x^{2}}{72} - \frac{x}{6} + \frac{11}{18}$ .

$y = \frac{x^{2}}{72} - \frac{x}{6} + \frac{11}{18}$