Find the Function Rule table[[x,f(x)],[9,2],[27,3],[81,4]]

$\begin{array}{c} x & f (x) \\ 9 & 2 \\ 27 & 3 \\ 81 & 4 \end{array}$

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 $f (x) = a x + b$ .

$2 = a (9) + b 3 = a (27) + b 4 = a (81) + b$

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

Solve for $b$ in $2 = a (9) + b$ .

Rewrite the equation as $a (9) + b = 2$ .

$a (9) + b = 2$

$3 = a (27) + b$

$4 = a (81) + b$

Move $9$ to the left of $a$ .

$9 a + b = 2$

$3 = a (27) + b$

$4 = a (81) + b$

Subtract $9 a$ from both sides of the equation.

$b = 2 - 9 a$

$3 = a (27) + b$

$4 = a (81) + b$

$b = 2 - 9 a$

$3 = a (27) + b$

$4 = a (81) + b$

Replace all occurrences of $b$ with $2 - 9 a$ in each equation.

Replace all occurrences of $b$ in $3 = a (27) + b$ with $2 - 9 a$ .

$3 = a (27) + 2 - 9 a$

$b = 2 - 9 a$

$4 = a (81) + b$

Simplify $3 = a (27) + 2 - 9 a$ .

Simplify the left side.

Remove parentheses.

$3 = a (27) + 2 - 9 a$

$b = 2 - 9 a$

$4 = a (81) + b$

$3 = a (27) + 2 - 9 a$

$b = 2 - 9 a$

$4 = a (81) + b$

Simplify the right side.

Simplify $a (27) + 2 - 9 a$ .

Move $27$ to the left of $a$ .

$3 = 27 a + 2 - 9 a$

$b = 2 - 9 a$

$4 = a (81) + b$

Subtract $9 a$ from $27 a$ .

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = a (81) + b$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = a (81) + b$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = a (81) + b$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = a (81) + b$

Replace all occurrences of $b$ in $4 = a (81) + b$ with $2 - 9 a$ .

$4 = a (81) + 2 - 9 a$

$3 = 18 a + 2$

$b = 2 - 9 a$

Simplify $4 = a (81) + 2 - 9 a$ .

Simplify the left side.

Remove parentheses.

$4 = a (81) + 2 - 9 a$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = a (81) + 2 - 9 a$

$3 = 18 a + 2$

$b = 2 - 9 a$

Simplify the right side.

Simplify $a (81) + 2 - 9 a$ .

Move $81$ to the left of $a$ .

$4 = 81 a + 2 - 9 a$

$3 = 18 a + 2$

$b = 2 - 9 a$

Subtract $9 a$ from $81 a$ .

$4 = 72 a + 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = 72 a + 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = 72 a + 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = 72 a + 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

$4 = 72 a + 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

Solve for $a$ in $4 = 72 a + 2$ .

Rewrite the equation as $72 a + 2 = 4$ .

$72 a + 2 = 4$

$3 = 18 a + 2$

$b = 2 - 9 a$

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

Subtract $2$ from both sides of the equation.

$72 a = 4 - 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

Subtract $2$ from $4$ .

$72 a = 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

$72 a = 2$

$3 = 18 a + 2$

$b = 2 - 9 a$

Divide each term in $72 a = 2$ by $72$ and simplify.

Divide each term in $72 a = 2$ by $72$ .

$\frac{72 a}{72} = \frac{2}{72}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Simplify the left side.

Cancel the common factor of $72$ .

Cancel the common factor.

$\frac{72 a}{72} = \frac{2}{72}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Divide $a$ by $1$ .

$a = \frac{2}{72}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{2}{72}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{2}{72}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Simplify the right side.

Cancel the common factor of $2$ and $72$ .

Factor $2$ out of $2$ .

$a = \frac{2 (1)}{72}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Cancel the common factors.

Factor $2$ out of $72$ .

$a = \frac{2 \cdot 1}{2 \cdot 36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Cancel the common factor.

$a = \frac{2 \cdot 1}{2 \cdot 36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Rewrite the expression.

$a = \frac{1}{36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{1}{36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{1}{36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{1}{36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{1}{36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

$a = \frac{1}{36}$

$3 = 18 a + 2$

$b = 2 - 9 a$

Replace all occurrences of $a$ with $\frac{1}{36}$ in each equation.

Replace all occurrences of $a$ in $3 = 18 a + 2$ with $\frac{1}{36}$ .

$3 = 18 (\frac{1}{36}) + 2$

$a = \frac{1}{36}$

$b = 2 - 9 a$

Simplify the right side.

Simplify $18 (\frac{1}{36}) + 2$ .

Cancel the common factor of $18$ .

Factor $18$ out of $36$ .

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

$a = \frac{1}{36}$

$b = 2 - 9 a$

Cancel the common factor.

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

$a = \frac{1}{36}$

$b = 2 - 9 a$

Rewrite the expression.

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

$a = \frac{1}{36}$

$b = 2 - 9 a$

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

$a = \frac{1}{36}$

$b = 2 - 9 a$

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

$3 = \frac{1}{2} + 2 \cdot \frac{2}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

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

$3 = \frac{1}{2} + \frac{2 \cdot 2}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

Combine the numerators over the common denominator.

$3 = \frac{1 + 2 \cdot 2}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

Simplify the numerator.

Multiply $2$ by $2$ .

$3 = \frac{1 + 4}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

Add $1$ and $4$ .

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = 2 - 9 a$

Replace all occurrences of $a$ in $b = 2 - 9 a$ with $\frac{1}{36}$ .

$b = 2 - 9 (\frac{1}{36})$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Simplify the right side.

Simplify $2 - 9 (\frac{1}{36})$ .

Simplify each term.

Cancel the common factor of $9$ .

Factor $9$ out of $- 9$ .

$b = 2 + 9 (- 1) (\frac{1}{36})$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Factor $9$ out of $36$ .

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

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Cancel the common factor.

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

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Rewrite the expression.

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

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

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

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Rewrite $- 1 (\frac{1}{4})$ as $- (\frac{1}{4})$ .

$b = 2 - \frac{1}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = 2 - \frac{1}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$b = 2 \cdot \frac{4}{4} - \frac{1}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Combine $2$ and $\frac{4}{4}$ .

$b = \frac{2 \cdot 4}{4} - \frac{1}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Combine the numerators over the common denominator.

$b = \frac{2 \cdot 4 - 1}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Simplify the numerator.

Multiply $2$ by $4$ .

$b = \frac{8 - 1}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Subtract $1$ from $8$ .

$b = \frac{7}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = \frac{7}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = \frac{7}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = \frac{7}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

$b = \frac{7}{4}$

$3 = \frac{5}{2}$

$a = \frac{1}{36}$

Since $3 = \frac{5}{2}$ is not true, there is no solution.

No solution

Since $y \neq f (x)$ 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 $f (x) = a x^{2} + b x + c$ .

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

Solve for $c$ in $2 = a \cdot 9^{2} + b (9) + c$ .

Rewrite the equation as $a \cdot 9^{2} + b (9) + c = 2$ .

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

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

Simplify each term.

Raise $9$ to the power of $2$ .

$a \cdot 81 + b (9) + c = 2$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

Move $81$ to the left of $a$ .

$81 \cdot a + b (9) + c = 2$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

Move $9$ to the left of $b$ .

$81 a + 9 b + c = 2$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

$81 a + 9 b + c = 2$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

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

Subtract $81 a$ from both sides of the equation.

$9 b + c = 2 - 81 a$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

Subtract $9 b$ from both sides of the equation.

$c = 2 - 81 a - 9 b$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

$c = 2 - 81 a - 9 b$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

$c = 2 - 81 a - 9 b$

$3 = a \cdot 27^{2} + b (27) + c$

$4 = a \cdot 81^{2} + b (81) + c$

Replace all occurrences of $c$ with $2 - 81 a - 9 b$ in each equation.

Replace all occurrences of $c$ in $3 = a \cdot 27^{2} + b (27) + c$ with $2 - 81 a - 9 b$ .

$3 = a \cdot 27^{2} + b (27) + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Simplify $3 = a \cdot 27^{2} + b (27) + 2 - 81 a - 9 b$ .

Simplify the left side.

Remove parentheses.

$3 = a \cdot 27^{2} + b (27) + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

$3 = a \cdot 27^{2} + b (27) + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Simplify the right side.

Simplify $a \cdot 27^{2} + b (27) + 2 - 81 a - 9 b$ .

Simplify each term.

Raise $27$ to the power of $2$ .

$3 = a \cdot 729 + b (27) + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Move $729$ to the left of $a$ .

$3 = 729 \cdot a + b (27) + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Move $27$ to the left of $b$ .

$3 = 729 a + 27 b + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

$3 = 729 a + 27 b + 2 - 81 a - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Simplify by adding terms.

Subtract $81 a$ from $729 a$ .

$3 = 648 a + 27 b + 2 - 9 b$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Subtract $9 b$ from $27 b$ .

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + c$

Replace all occurrences of $c$ in $4 = a \cdot 81^{2} + b (81) + c$ with $2 - 81 a - 9 b$ .

$4 = a \cdot 81^{2} + b (81) + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Simplify $4 = a \cdot 81^{2} + b (81) + 2 - 81 a - 9 b$ .

Simplify the left side.

Remove parentheses.

$4 = a \cdot 81^{2} + b (81) + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = a \cdot 81^{2} + b (81) + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Simplify the right side.

Simplify $a \cdot 81^{2} + b (81) + 2 - 81 a - 9 b$ .

Simplify each term.

Raise $81$ to the power of $2$ .

$4 = a \cdot 6561 + b (81) + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Move $6561$ to the left of $a$ .

$4 = 6561 \cdot a + b (81) + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Move $81$ to the left of $b$ .

$4 = 6561 a + 81 b + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = 6561 a + 81 b + 2 - 81 a - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Simplify by adding terms.

Subtract $81 a$ from $6561 a$ .

$4 = 6480 a + 81 b + 2 - 9 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Subtract $9 b$ from $81 b$ .

$4 = 6480 a + 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = 6480 a + 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = 6480 a + 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = 6480 a + 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = 6480 a + 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$4 = 6480 a + 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Solve for $a$ in $4 = 6480 a + 72 b + 2$ .

Rewrite the equation as $6480 a + 72 b + 2 = 4$ .

$6480 a + 72 b + 2 = 4$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

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

Subtract $72 b$ from both sides of the equation.

$6480 a + 2 = 4 - 72 b$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Subtract $2$ from both sides of the equation.

$6480 a = 4 - 72 b - 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Subtract $2$ from $4$ .

$6480 a = - 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$6480 a = - 72 b + 2$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Divide each term in $6480 a = - 72 b + 2$ by $6480$ and simplify.

Divide each term in $6480 a = - 72 b + 2$ by $6480$ .

$\frac{6480 a}{6480} = \frac{- 72 b}{6480} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Simplify the left side.

Cancel the common factor of $6480$ .

Cancel the common factor.

$\frac{6480 a}{6480} = \frac{- 72 b}{6480} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Divide $a$ by $1$ .

$a = \frac{- 72 b}{6480} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = \frac{- 72 b}{6480} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = \frac{- 72 b}{6480} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 72$ and $6480$ .

Factor $72$ out of $- 72 b$ .

$a = \frac{72 (- b)}{6480} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Cancel the common factors.

Factor $72$ out of $6480$ .

$a = \frac{72 (- b)}{72 (90)} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Cancel the common factor.

$a = \frac{72 (- b)}{72 \cdot 90} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Rewrite the expression.

$a = \frac{- b}{90} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = \frac{- b}{90} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = \frac{- b}{90} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Move the negative in front of the fraction.

$a = - \frac{b}{90} + \frac{2}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Cancel the common factor of $2$ and $6480$ .

Factor $2$ out of $2$ .

$a = - \frac{b}{90} + \frac{2 (1)}{6480}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Cancel the common factors.

Factor $2$ out of $6480$ .

$a = - \frac{b}{90} + \frac{2 \cdot 1}{2 \cdot 3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Cancel the common factor.

$a = - \frac{b}{90} + \frac{2 \cdot 1}{2 \cdot 3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Rewrite the expression.

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

$a = - \frac{b}{90} + \frac{1}{3240}$

$3 = 648 a + 18 b + 2$

$c = 2 - 81 a - 9 b$

Replace all occurrences of $a$ with $- \frac{b}{90} + \frac{1}{3240}$ in each equation.

Replace all occurrences of $a$ in $3 = 648 a + 18 b + 2$ with $- \frac{b}{90} + \frac{1}{3240}$ .

$3 = 648 (- \frac{b}{90} + \frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Simplify the right side.

Simplify $648 (- \frac{b}{90} + \frac{1}{3240}) + 18 b + 2$ .

Simplify each term.

Apply the distributive property.

$3 = 648 (- \frac{b}{90}) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Cancel the common factor of $18$ .

Move the leading negative in $- \frac{b}{90}$ into the numerator.

$3 = 648 (\frac{- b}{90}) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Factor $18$ out of $648$ .

$3 = 18 (36) (\frac{- b}{90}) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Factor $18$ out of $90$ .

$3 = 18 \cdot (36 (\frac{- b}{18 \cdot 5})) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Cancel the common factor.

$3 = 18 \cdot (36 (\frac{- b}{18 \cdot 5})) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Rewrite the expression.

$3 = 36 (\frac{- b}{5}) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = 36 (\frac{- b}{5}) + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Combine $36$ and $\frac{- b}{5}$ .

$3 = \frac{36 (- b)}{5} + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Multiply $- 1$ by $36$ .

$3 = \frac{- 36 b}{5} + 648 (\frac{1}{3240}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Cancel the common factor of $648$ .

Factor $648$ out of $3240$ .

$3 = \frac{- 36 b}{5} + 648 (\frac{1}{648 (5)}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Cancel the common factor.

$3 = \frac{- 36 b}{5} + 648 (\frac{1}{648 \cdot 5}) + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Rewrite the expression.

$3 = \frac{- 36 b}{5} + \frac{1}{5} + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = \frac{- 36 b}{5} + \frac{1}{5} + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Move the negative in front of the fraction.

$3 = - \frac{36 b}{5} + \frac{1}{5} + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = - \frac{36 b}{5} + \frac{1}{5} + 18 b + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

To write $18 b$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$3 = - \frac{36 b}{5} + 18 b \cdot \frac{5}{5} + \frac{1}{5} + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Combine $18 b$ and $\frac{5}{5}$ .

$3 = - \frac{36 b}{5} + \frac{18 b \cdot 5}{5} + \frac{1}{5} + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Combine the numerators over the common denominator.

$3 = \frac{- 36 b + 18 b \cdot 5}{5} + \frac{1}{5} + 2$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Combine the numerators over the common denominator.

$3 = 2 + \frac{- 36 b + 18 b \cdot 5 + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Multiply $5$ by $18$ .

$3 = 2 + \frac{- 36 b + 90 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Add $- 36 b$ and $90 b$ .

$3 = 2 + \frac{54 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

To write $2$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$3 = 2 \cdot \frac{5}{5} + \frac{54 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Simplify terms.

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

$3 = \frac{2 \cdot 5}{5} + \frac{54 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Combine the numerators over the common denominator.

$3 = \frac{2 \cdot 5 + 54 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = \frac{2 \cdot 5 + 54 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Simplify the numerator.

Multiply $2$ by $5$ .

$3 = \frac{10 + 54 b + 1}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Add $10$ and $1$ .

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 81 a - 9 b$

Replace all occurrences of $a$ in $c = 2 - 81 a - 9 b$ with $- \frac{b}{90} + \frac{1}{3240}$ .

$c = 2 - 81 (- \frac{b}{90} + \frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the right side.

Simplify $2 - 81 (- \frac{b}{90} + \frac{1}{3240}) - 9 b$ .

Simplify each term.

Apply the distributive property.

$c = 2 - 81 (- \frac{b}{90}) - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor of $9$ .

Move the leading negative in $- \frac{b}{90}$ into the numerator.

$c = 2 - 81 \frac{- b}{90} - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Factor $9$ out of $- 81$ .

$c = 2 + 9 (- 9) (\frac{- b}{90}) - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Factor $9$ out of $90$ .

$c = 2 + 9 \cdot (- 9 \frac{- b}{9 \cdot 10}) - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor.

$c = 2 + 9 \cdot (- 9 \frac{- b}{9 \cdot 10}) - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite the expression.

$c = 2 - 9 \frac{- b}{10} - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 - 9 \frac{- b}{10} - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Combine $- 9$ and $\frac{- b}{10}$ .

$c = 2 + \frac{- 9 (- b)}{10} - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Multiply $- 1$ by $- 9$ .

$c = 2 + \frac{9 b}{10} - 81 (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor of $81$ .

Factor $81$ out of $- 81$ .

$c = 2 + \frac{9 b}{10} + 81 (- 1) (\frac{1}{3240}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Factor $81$ out of $3240$ .

$c = 2 + \frac{9 b}{10} + 81 \cdot (- 1 \frac{1}{81 \cdot 40}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor.

$c = 2 + \frac{9 b}{10} + 81 \cdot (- 1 \frac{1}{81 \cdot 40}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite the expression.

$c = 2 + \frac{9 b}{10} - 1 (\frac{1}{40}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 + \frac{9 b}{10} - 1 (\frac{1}{40}) - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite $- 1 (\frac{1}{40})$ as $- (\frac{1}{40})$ .

$c = 2 + \frac{9 b}{10} - \frac{1}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = 2 + \frac{9 b}{10} - \frac{1}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{40}{40}$ .

$c = \frac{9 b}{10} + 2 \cdot \frac{40}{40} - \frac{1}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Combine $2$ and $\frac{40}{40}$ .

$c = \frac{9 b}{10} + \frac{2 \cdot 40}{40} - \frac{1}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Combine the numerators over the common denominator.

$c = \frac{9 b}{10} + \frac{2 \cdot 40 - 1}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the numerator.

Multiply $2$ by $40$ .

$c = \frac{9 b}{10} + \frac{80 - 1}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Subtract $1$ from $80$ .

$c = \frac{9 b}{10} + \frac{79}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{9 b}{10} + \frac{79}{40} - 9 b$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

To write $- 9 b$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$c = \frac{9 b}{10} - 9 b \cdot \frac{10}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify terms.

Combine $- 9 b$ and $\frac{10}{10}$ .

$c = \frac{9 b}{10} + \frac{- 9 b \cdot 10}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Combine the numerators over the common denominator.

$c = \frac{9 b - 9 b \cdot 10}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{9 b - 9 b \cdot 10}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify each term.

Simplify the numerator.

Factor $9 b$ out of $9 b - 9 b \cdot 10$ .

Factor $9 b$ out of $9 b$ .

$c = \frac{9 b (1) - 9 b \cdot 10}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Factor $9 b$ out of $- 9 b \cdot 10$ .

$c = \frac{9 b (1) + 9 b (- 1 \cdot 10)}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Factor $9 b$ out of $9 b (1) + 9 b (- 1 \cdot 10)$ .

$c = \frac{9 b (1 - 1 \cdot 10)}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{9 b (1 - 1 \cdot 10)}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Multiply $- 1$ by $10$ .

$c = \frac{9 b (1 - 10)}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Subtract $10$ from $1$ .

$c = \frac{9 b \cdot - 9}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Multiply $- 9$ by $9$ .

$c = \frac{- 81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{- 81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Move the negative in front of the fraction.

$c = - \frac{81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$3 = \frac{54 b + 11}{5}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Solve for $b$ in $3 = \frac{54 b + 11}{5}$ .

Rewrite the equation as $\frac{54 b + 11}{5} = 3$ .

$\frac{54 b + 11}{5} = 3$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Multiply both sides by $5$ .

$\frac{54 b + 11}{5} \cdot 5 = 3 \cdot 5$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify.

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{54 b + 11}{5} \cdot 5 = 3 \cdot 5$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite the expression.

$54 b + 11 = 3 \cdot 5$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$54 b + 11 = 3 \cdot 5$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$54 b + 11 = 3 \cdot 5$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the right side.

Multiply $3$ by $5$ .

$54 b + 11 = 15$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$54 b + 11 = 15$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$54 b + 11 = 15$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Solve for $b$ .

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

Subtract $11$ from both sides of the equation.

$54 b = 15 - 11$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Subtract $11$ from $15$ .

$54 b = 4$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$54 b = 4$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Divide each term in $54 b = 4$ by $54$ and simplify.

Divide each term in $54 b = 4$ by $54$ .

$\frac{54 b}{54} = \frac{4}{54}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the left side.

Cancel the common factor of $54$ .

Cancel the common factor.

$\frac{54 b}{54} = \frac{4}{54}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Divide $b$ by $1$ .

$b = \frac{4}{54}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{4}{54}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{4}{54}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the right side.

Cancel the common factor of $4$ and $54$ .

Factor $2$ out of $4$ .

$b = \frac{2 (2)}{54}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factors.

Factor $2$ out of $54$ .

$b = \frac{2 \cdot 2}{2 \cdot 27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor.

$b = \frac{2 \cdot 2}{2 \cdot 27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite the expression.

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$b = \frac{2}{27}$

$c = - \frac{81 b}{10} + \frac{79}{40}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Replace all occurrences of $b$ with $\frac{2}{27}$ in each equation.

Replace all occurrences of $b$ in $c = - \frac{81 b}{10} + \frac{79}{40}$ with $\frac{2}{27}$ .

$c = - \frac{81 (\frac{2}{27})}{10} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the right side.

Simplify $- \frac{81 (\frac{2}{27})}{10} + \frac{79}{40}$ .

Simplify each term.

Combine $81$ and $\frac{2}{27}$ .

$c = - \frac{\frac{81 \cdot 2}{27}}{10} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Multiply $81$ by $2$ .

$c = - \frac{\frac{162}{27}}{10} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Divide $162$ by $27$ .

$c = - \frac{6}{10} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor of $6$ and $10$ .

Factor $2$ out of $6$ .

$c = - \frac{2 (3)}{10} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factors.

Factor $2$ out of $10$ .

$c = - \frac{2 \cdot 3}{2 \cdot 5} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor.

$c = - \frac{2 \cdot 3}{2 \cdot 5} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite the expression.

$c = - \frac{3}{5} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{3}{5} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{3}{5} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{3}{5} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

To write $- \frac{3}{5}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$c = - \frac{3}{5} \cdot \frac{8}{8} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Write each expression with a common denominator of $40$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{5}$ by $\frac{8}{8}$ .

$c = - \frac{3 \cdot 8}{5 \cdot 8} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Multiply $5$ by $8$ .

$c = - \frac{3 \cdot 8}{40} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = - \frac{3 \cdot 8}{40} + \frac{79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Combine the numerators over the common denominator.

$c = \frac{- 3 \cdot 8 + 79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Simplify the numerator.

Multiply $- 3$ by $8$ .

$c = \frac{- 24 + 79}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Add $- 24$ and $79$ .

$c = \frac{55}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{55}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor of $55$ and $40$ .

Factor $5$ out of $55$ .

$c = \frac{5 (11)}{40}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factors.

Factor $5$ out of $40$ .

$c = \frac{5 \cdot 11}{5 \cdot 8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Cancel the common factor.

$c = \frac{5 \cdot 11}{5 \cdot 8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Rewrite the expression.

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{b}{90} + \frac{1}{3240}$

Replace all occurrences of $b$ in $a = - \frac{b}{90} + \frac{1}{3240}$ with $\frac{2}{27}$ .

$a = - \frac{\frac{2}{27}}{90} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Simplify the right side.

Simplify $- \frac{\frac{2}{27}}{90} + \frac{1}{3240}$ .

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$a = - (\frac{2}{27} \cdot \frac{1}{90}) + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Cancel the common factor of $2$ .

Factor $2$ out of $90$ .

$a = - (\frac{2}{27} \cdot \frac{1}{2 (45)}) + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Cancel the common factor.

$a = - (\frac{2}{27} \cdot \frac{1}{2 \cdot 45}) + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Rewrite the expression.

$a = - (\frac{1}{27} \cdot \frac{1}{45}) + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - (\frac{1}{27} \cdot \frac{1}{45}) + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Multiply $\frac{1}{27}$ by $\frac{1}{45}$ .

$a = - \frac{1}{27 \cdot 45} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Multiply $27$ by $45$ .

$a = - \frac{1}{1215} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{1}{1215} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

To write $- \frac{1}{1215}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$a = - \frac{1}{1215} \cdot \frac{8}{8} + \frac{1}{3240}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

To write $\frac{1}{3240}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$a = - \frac{1}{1215} \cdot \frac{8}{8} + \frac{1}{3240} \cdot \frac{3}{3}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Write each expression with a common denominator of $9720$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{1215}$ by $\frac{8}{8}$ .

$a = - \frac{8}{1215 \cdot 8} + \frac{1}{3240} \cdot \frac{3}{3}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Multiply $1215$ by $8$ .

$a = - \frac{8}{9720} + \frac{1}{3240} \cdot \frac{3}{3}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Multiply $\frac{1}{3240}$ by $\frac{3}{3}$ .

$a = - \frac{8}{9720} + \frac{3}{3240 \cdot 3}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Multiply $3240$ by $3$ .

$a = - \frac{8}{9720} + \frac{3}{9720}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{8}{9720} + \frac{3}{9720}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Simplify the expression.

Combine the numerators over the common denominator.

$a = \frac{- 8 + 3}{9720}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Add $- 8$ and $3$ .

$a = \frac{- 5}{9720}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = \frac{- 5}{9720}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Cancel the common factor of $- 5$ and $9720$ .

Factor $5$ out of $- 5$ .

$a = \frac{5 (- 1)}{9720}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Cancel the common factors.

Factor $5$ out of $9720$ .

$a = \frac{5 \cdot - 1}{5 \cdot 1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Cancel the common factor.

$a = \frac{5 \cdot - 1}{5 \cdot 1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Rewrite the expression.

$a = \frac{- 1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = \frac{- 1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = \frac{- 1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

Move the negative in front of the fraction.

$a = - \frac{1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

$a = - \frac{1}{1944}$

$c = \frac{11}{8}$

$b = \frac{2}{27}$

List all of the solutions.

$a = - \frac{1}{1944}, c = \frac{11}{8}, b = \frac{2}{27}$

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

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = - \frac{1}{1944}$ , $b = \frac{2}{27}$ , $c = \frac{11}{8}$ , and $x = 9$ .

Simplify each term.

Raise $9$ to the power of $2$ .

$y = - \frac{1}{1944} \cdot 81 + (\frac{2}{27}) \cdot (9) + \frac{11}{8}$

Cancel the common factor of $81$ .

Move the leading negative in $- \frac{1}{1944}$ into the numerator.

$y = \frac{- 1}{1944} \cdot 81 + (\frac{2}{27}) \cdot (9) + \frac{11}{8}$

Factor $81$ out of $1944$ .

$y = \frac{- 1}{81 (24)} \cdot 81 + (\frac{2}{27}) \cdot (9) + \frac{11}{8}$

Cancel the common factor.

$y = \frac{- 1}{81 \cdot 24} \cdot 81 + (\frac{2}{27}) \cdot (9) + \frac{11}{8}$

Rewrite the expression.

$y = \frac{- 1}{24} + (\frac{2}{27}) \cdot (9) + \frac{11}{8}$

Move the negative in front of the fraction.

$y = - \frac{1}{24} + (\frac{2}{27}) \cdot (9) + \frac{11}{8}$

Cancel the common factor of $9$ .

Factor $9$ out of $27$ .

$y = - \frac{1}{24} + \frac{2}{9 (3)} \cdot 9 + \frac{11}{8}$

Cancel the common factor.

$y = - \frac{1}{24} + \frac{2}{9 \cdot 3} \cdot 9 + \frac{11}{8}$

Rewrite the expression.

$y = - \frac{1}{24} + \frac{2}{3} + \frac{11}{8}$

Find the common denominator.

Multiply $\frac{2}{3}$ by $\frac{8}{8}$ .

$y = - \frac{1}{24} + \frac{2}{3} \cdot \frac{8}{8} + \frac{11}{8}$

Multiply $\frac{2}{3}$ by $\frac{8}{8}$ .

$y = - \frac{1}{24} + \frac{2 \cdot 8}{3 \cdot 8} + \frac{11}{8}$

Multiply $\frac{11}{8}$ by $\frac{3}{3}$ .

$y = - \frac{1}{24} + \frac{2 \cdot 8}{3 \cdot 8} + \frac{11}{8} \cdot \frac{3}{3}$

Multiply $\frac{11}{8}$ by $\frac{3}{3}$ .

$y = - \frac{1}{24} + \frac{2 \cdot 8}{3 \cdot 8} + \frac{11 \cdot 3}{8 \cdot 3}$

Multiply $3$ by $8$ .

$y = - \frac{1}{24} + \frac{2 \cdot 8}{24} + \frac{11 \cdot 3}{8 \cdot 3}$

Reorder the factors of $8 \cdot 3$ .

$y = - \frac{1}{24} + \frac{2 \cdot 8}{24} + \frac{11 \cdot 3}{3 \cdot 8}$

Multiply $3$ by $8$ .

$y = - \frac{1}{24} + \frac{2 \cdot 8}{24} + \frac{11 \cdot 3}{24}$

Combine the numerators over the common denominator.

$y = \frac{- 1 + 2 \cdot 8 + 11 \cdot 3}{24}$

Simplify each term.

Multiply $2$ by $8$ .

$y = \frac{- 1 + 16 + 11 \cdot 3}{24}$

Multiply $11$ by $3$ .

$y = \frac{- 1 + 16 + 33}{24}$

Simplify the expression.

Add $- 1$ and $16$ .

$y = \frac{15 + 33}{24}$

Add $15$ and $33$ .

$y = \frac{48}{24}$

Divide $48$ by $24$ .

$y = 2$

If the table has a quadratic function rule, $y = f (x)$ for the corresponding $x$ value, $x = 9$ . This check passes since $y = 2$ and $f (x) = 2$ .

$2 = 2$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = - \frac{1}{1944}$ , $b = \frac{2}{27}$ , $c = \frac{11}{8}$ , and $x = 27$ .

Simplify each term.

Raise $27$ to the power of $2$ .

$y = - \frac{1}{1944} \cdot 729 + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Cancel the common factor of $243$ .

Move the leading negative in $- \frac{1}{1944}$ into the numerator.

$y = \frac{- 1}{1944} \cdot 729 + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Factor $243$ out of $1944$ .

$y = \frac{- 1}{243 (8)} \cdot 729 + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Factor $243$ out of $729$ .

$y = \frac{- 1}{243 \cdot 8} \cdot (243 \cdot 3) + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Cancel the common factor.

$y = \frac{- 1}{243 \cdot 8} \cdot (243 \cdot 3) + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Rewrite the expression.

$y = \frac{- 1}{8} \cdot 3 + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Combine $\frac{- 1}{8}$ and $3$ .

$y = \frac{- 1 \cdot 3}{8} + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Multiply $- 1$ by $3$ .

$y = \frac{- 3}{8} + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Move the negative in front of the fraction.

$y = - \frac{3}{8} + (\frac{2}{27}) \cdot (27) + \frac{11}{8}$

Cancel the common factor of $27$ .

Cancel the common factor.

$y = - \frac{3}{8} + \frac{2}{27} \cdot 27 + \frac{11}{8}$

Rewrite the expression.

$y = - \frac{3}{8} + 2 + \frac{11}{8}$

Combine fractions.

Combine the numerators over the common denominator.

$y = 2 + \frac{- 3 + 11}{8}$

Simplify the expression.

Add $- 3$ and $11$ .

$y = 2 + \frac{8}{8}$

Divide $8$ by $8$ .

$y = 2 + 1$

Add $2$ and $1$ .

$y = 3$

If the table has a quadratic function rule, $y = f (x)$ for the corresponding $x$ value, $x = 27$ . This check passes since $y = 3$ and $f (x) = 3$ .

$3 = 3$

Calculate the value of $y$ such that $y = a x^{2} + b$ when $a = - \frac{1}{1944}$ , $b = \frac{2}{27}$ , $c = \frac{11}{8}$ , and $x = 81$ .

Simplify each term.

Raise $81$ to the power of $2$ .

$y = - \frac{1}{1944} \cdot 6561 + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Cancel the common factor of $243$ .

Move the leading negative in $- \frac{1}{1944}$ into the numerator.

$y = \frac{- 1}{1944} \cdot 6561 + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Factor $243$ out of $1944$ .

$y = \frac{- 1}{243 (8)} \cdot 6561 + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Factor $243$ out of $6561$ .

$y = \frac{- 1}{243 \cdot 8} \cdot (243 \cdot 27) + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Cancel the common factor.

$y = \frac{- 1}{243 \cdot 8} \cdot (243 \cdot 27) + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Rewrite the expression.

$y = \frac{- 1}{8} \cdot 27 + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Combine $\frac{- 1}{8}$ and $27$ .

$y = \frac{- 1 \cdot 27}{8} + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Multiply $- 1$ by $27$ .

$y = \frac{- 27}{8} + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Move the negative in front of the fraction.

$y = - \frac{27}{8} + (\frac{2}{27}) \cdot (81) + \frac{11}{8}$

Cancel the common factor of $27$ .

Factor $27$ out of $81$ .

$y = - \frac{27}{8} + \frac{2}{27} \cdot (27 (3)) + \frac{11}{8}$

Cancel the common factor.

$y = - \frac{27}{8} + \frac{2}{27} \cdot (27 \cdot 3) + \frac{11}{8}$

Rewrite the expression.

$y = - \frac{27}{8} + 2 \cdot 3 + \frac{11}{8}$

Multiply $2$ by $3$ .

$y = - \frac{27}{8} + 6 + \frac{11}{8}$

Combine fractions.

Combine the numerators over the common denominator.

$y = 6 + \frac{- 27 + 11}{8}$

Simplify the expression.

Add $- 27$ and $11$ .

$y = 6 + \frac{- 16}{8}$

Divide $- 16$ by $8$ .

$y = 6 - 2$

Subtract $2$ from $6$ .

$y = 4$

If the table has a quadratic function rule, $y = f (x)$ for the corresponding $x$ value, $x = 81$ . This check passes since $y = 4$ and $f (x) = 4$ .

$4 = 4$

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

The function is quadratic

Step 3

Since all $y = f (x)$ , the function is quadratic and follows the form $y = - \frac{x^{2}}{1944} + \frac{2 x}{27} + \frac{11}{8}$ .

$y = - \frac{x^{2}}{1944} + \frac{2 x}{27} + \frac{11}{8}$

Find the Function Rule table[[x,f(x)],[9,2],[27,3],[81,4]]

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