Basic Math Examples

$\frac{9 y^{2} - 28 + 3}{9 y^{2} + 80 y - 9} \div \frac{27 y^{2} - 12 y + 1}{3 y^{2} + 26 y - 9}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{9 y^{2} - 28 + 3}{9 y^{2} + 80 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 2

Simplify the numerator.

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Step 2.1

Add $- 28$ and $3$ .

$\frac{9 y^{2} - 25}{9 y^{2} + 80 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 2.2

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

$\frac{{(3 y)}^{2} - 25}{9 y^{2} + 80 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 2.3

Rewrite $25$ as $5^{2}$ .

$\frac{{(3 y)}^{2} - 5^{2}}{9 y^{2} + 80 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 2.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 y$ and $b = 5$ .

$\frac{(3 y + 5) (3 y - 5)}{9 y^{2} + 80 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 3

Factor by grouping.

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Step 3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot - 9 = - 81$ and whose sum is $b = 80$ .

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Step 3.1.1

Factor $80$ out of $80 y$ .

$\frac{(3 y + 5) (3 y - 5)}{9 y^{2} + 80 (y) - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 3.1.2

Rewrite $80$ as $- 1$ plus $81$

$\frac{(3 y + 5) (3 y - 5)}{9 y^{2} + (- 1 + 81) y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 3.1.3

Apply the distributive property.

$\frac{(3 y + 5) (3 y - 5)}{9 y^{2} - 1 y + 81 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 3.2

Factor out the greatest common factor from each group.

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Step 3.2.1

Group the first two terms and the last two terms.

$\frac{(3 y + 5) (3 y - 5)}{(9 y^{2} - 1 y) + 81 y - 9} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 3.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(3 y + 5) (3 y - 5)}{y (9 y - 1) + 9 (9 y - 1)} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 3.3

Factor the polynomial by factoring out the greatest common factor, $9 y - 1$ .

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{3 y^{2} + 26 y - 9}{27 y^{2} - 12 y + 1}$

Step 4

Factor by grouping.

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Step 4.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 9 = - 27$ and whose sum is $b = 26$ .

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Step 4.1.1

Factor $26$ out of $26 y$ .

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{3 y^{2} + 26 (y) - 9}{27 y^{2} - 12 y + 1}$

Step 4.1.2

Rewrite $26$ as $- 1$ plus $27$

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{3 y^{2} + (- 1 + 27) y - 9}{27 y^{2} - 12 y + 1}$

Step 4.1.3

Apply the distributive property.

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{3 y^{2} - 1 y + 27 y - 9}{27 y^{2} - 12 y + 1}$

Step 4.2

Factor out the greatest common factor from each group.

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Step 4.2.1

Group the first two terms and the last two terms.

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y^{2} - 1 y) + 27 y - 9}{27 y^{2} - 12 y + 1}$

Step 4.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{y (3 y - 1) + 9 (3 y - 1)}{27 y^{2} - 12 y + 1}$

Step 4.3

Factor the polynomial by factoring out the greatest common factor, $3 y - 1$ .

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{27 y^{2} - 12 y + 1}$

Step 5

Factor by grouping.

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Step 5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 27 \cdot 1 = 27$ and whose sum is $b = - 12$ .

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Step 5.1.1

Factor $- 12$ out of $- 12 y$ .

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{27 y^{2} - 12 (y) + 1}$

Step 5.1.2

Rewrite $- 12$ as $- 3$ plus $- 9$

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{27 y^{2} + (- 3 - 9) y + 1}$

Step 5.1.3

Apply the distributive property.

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{27 y^{2} - 3 y - 9 y + 1}$

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{(27 y^{2} - 3 y) - 9 y + 1}$

Step 5.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{3 y (9 y - 1) - (9 y - 1)}$

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $9 y - 1$ .

$\frac{(3 y + 5) (3 y - 5)}{(9 y - 1) (y + 9)} \cdot \frac{(3 y - 1) (y + 9)}{(9 y - 1) (3 y - 1)}$

Step 6

Cancel the common factor of $y + 9$ .

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Step 6.1

Factor $y + 9$ out of $(9 y - 1) (y + 9)$ .

$\frac{(3 y + 5) (3 y - 5)}{(y + 9) (9 y - 1)} \cdot \frac{(3 y - 1) (y + 9)}{(9 y - 1) (3 y - 1)}$

Step 6.2

Factor $y + 9$ out of $(3 y - 1) (y + 9)$ .

$\frac{(3 y + 5) (3 y - 5)}{(y + 9) (9 y - 1)} \cdot \frac{(y + 9) (3 y - 1)}{(9 y - 1) (3 y - 1)}$

Step 6.3

Cancel the common factor.

$\frac{(3 y + 5) (3 y - 5)}{(y + 9) (9 y - 1)} \cdot \frac{(y + 9) (3 y - 1)}{(9 y - 1) (3 y - 1)}$

Step 6.4

Rewrite the expression.

$\frac{(3 y + 5) (3 y - 5)}{9 y - 1} \cdot \frac{3 y - 1}{(9 y - 1) (3 y - 1)}$

Step 7

Multiply $\frac{(3 y + 5) (3 y - 5)}{9 y - 1}$ by $\frac{3 y - 1}{(9 y - 1) (3 y - 1)}$ .

$\frac{(3 y + 5) (3 y - 5) (3 y - 1)}{(9 y - 1) ((9 y - 1) (3 y - 1))}$

Step 8

Raise $9 y - 1$ to the power of $1$ .

$\frac{(3 y + 5) (3 y - 5) (3 y - 1)}{{(9 y - 1)}^{1} (9 y - 1) (3 y - 1)}$

Step 9

Raise $9 y - 1$ to the power of $1$ .

$\frac{(3 y + 5) (3 y - 5) (3 y - 1)}{{(9 y - 1)}^{1} {(9 y - 1)}^{1} (3 y - 1)}$

Step 10

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{(3 y + 5) (3 y - 5) (3 y - 1)}{{(9 y - 1)}^{1 + 1} (3 y - 1)}$

Step 11

Add $1$ and $1$ .

$\frac{(3 y + 5) (3 y - 5) (3 y - 1)}{{(9 y - 1)}^{2} (3 y - 1)}$

Step 12

Cancel the common factor of $3 y - 1$ .

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Step 12.1

Cancel the common factor.

$\frac{(3 y + 5) (3 y - 5) (3 y - 1)}{{(9 y - 1)}^{2} (3 y - 1)}$

Step 12.2

Rewrite the expression.

$\frac{(3 y + 5) (3 y - 5)}{{(9 y - 1)}^{2}}$