Algebra Examples

$\frac{x}{2 x - 5} - \frac{4 x + 15}{4 x^{2} - 25}$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\frac{x}{2 x - 5} - \frac{4 x + 15}{{(2 x)}^{2} - 25}$

Step 1.2

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

$\frac{x}{2 x - 5} - \frac{4 x + 15}{{(2 x)}^{2} - 5^{2}}$

Step 1.3

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

$\frac{x}{2 x - 5} - \frac{4 x + 15}{(2 x + 5) (2 x - 5)}$

Step 2

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

$\frac{x}{2 x - 5} \cdot \frac{2 x + 5}{2 x + 5} - \frac{4 x + 15}{(2 x + 5) (2 x - 5)}$

Step 3

Write each expression with a common denominator of $(2 x - 5) (2 x + 5)$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.1

Multiply $\frac{x}{2 x - 5}$ by $\frac{2 x + 5}{2 x + 5}$ .

$\frac{x (2 x + 5)}{(2 x - 5) (2 x + 5)} - \frac{4 x + 15}{(2 x + 5) (2 x - 5)}$

Step 3.2

Reorder the factors of $(2 x - 5) (2 x + 5)$ .

$\frac{x (2 x + 5)}{(2 x + 5) (2 x - 5)} - \frac{4 x + 15}{(2 x + 5) (2 x - 5)}$

Step 4

Combine the numerators over the common denominator.

$\frac{x (2 x + 5) - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

Step 5

Simplify the numerator.

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Step 5.1

Apply the distributive property.

$\frac{x (2 x) + x \cdot 5 - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

Step 5.2

Rewrite using the commutative property of multiplication.

$\frac{2 x \cdot x + x \cdot 5 - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

Step 5.3

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

$\frac{2 x \cdot x + 5 \cdot x - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

Step 5.4

Multiply $x$ by $x$ by adding the exponents.

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Step 5.4.1

Move $x$ .

$\frac{2 (x \cdot x) + 5 \cdot x - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

Step 5.4.2

Multiply $x$ by $x$ .

$\frac{2 x^{2} + 5 \cdot x - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

$\frac{2 x^{2} + 5 x - (4 x + 15)}{(2 x + 5) (2 x - 5)}$

Step 5.5

Apply the distributive property.

$\frac{2 x^{2} + 5 x - (4 x) - 1 \cdot 15}{(2 x + 5) (2 x - 5)}$

Step 5.6

Multiply $4$ by $- 1$ .

$\frac{2 x^{2} + 5 x - 4 x - 1 \cdot 15}{(2 x + 5) (2 x - 5)}$

Step 5.7

Multiply $- 1$ by $15$ .

$\frac{2 x^{2} + 5 x - 4 x - 15}{(2 x + 5) (2 x - 5)}$

Step 5.8

Subtract $4 x$ from $5 x$ .

$\frac{2 x^{2} + x - 15}{(2 x + 5) (2 x - 5)}$

Step 5.9

Factor by grouping.

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Step 5.9.1

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 $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = 1$ .

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Step 5.9.1.1

Multiply by $1$ .

$\frac{2 x^{2} + 1 x - 15}{(2 x + 5) (2 x - 5)}$

Step 5.9.1.2

Rewrite $1$ as $- 5$ plus $6$

$\frac{2 x^{2} + (- 5 + 6) x - 15}{(2 x + 5) (2 x - 5)}$

Step 5.9.1.3

Apply the distributive property.

$\frac{2 x^{2} - 5 x + 6 x - 15}{(2 x + 5) (2 x - 5)}$

Step 5.9.2

Factor out the greatest common factor from each group.

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Step 5.9.2.1

Group the first two terms and the last two terms.

$\frac{(2 x^{2} - 5 x) + 6 x - 15}{(2 x + 5) (2 x - 5)}$

Step 5.9.2.2

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

$\frac{x (2 x - 5) + 3 (2 x - 5)}{(2 x + 5) (2 x - 5)}$

Step 5.9.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$\frac{(2 x - 5) (x + 3)}{(2 x + 5) (2 x - 5)}$

Step 6

Cancel the common factor of $2 x - 5$ .

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Step 6.1

Cancel the common factor.

$\frac{(2 x - 5) (x + 3)}{(2 x + 5) (2 x - 5)}$

Step 6.2

Rewrite the expression.

$\frac{x + 3}{2 x + 5}$