Algebra Examples

$\frac{1}{x - 2} - \frac{2}{x + 5} = \frac{2 x - 1}{x^{2} + 3 x - 10}$

Step 1

Subtract $\frac{2 x - 1}{x^{2} + 3 x - 10}$ from both sides of the equation.

$\frac{1}{x - 2} - \frac{2}{x + 5} - \frac{2 x - 1}{x^{2} + 3 x - 10} = 0$

Step 2

Simplify $\frac{1}{x - 2} - \frac{2}{x + 5} - \frac{2 x - 1}{x^{2} + 3 x - 10}$ .

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Step 2.1

Factor $x^{2} + 3 x - 10$ using the AC method.

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Step 2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 10$ and whose sum is $3$ .

$- 2, 5$

Step 2.1.2

Write the factored form using these integers.

$\frac{1}{x - 2} - \frac{2}{x + 5} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.2

To write $\frac{1}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$\frac{1}{x - 2} \cdot \frac{x + 5}{x + 5} - \frac{2}{x + 5} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.3

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

$\frac{1}{x - 2} \cdot \frac{x + 5}{x + 5} - \frac{2}{x + 5} \cdot \frac{x - 2}{x - 2} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.4

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

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Step 2.4.1

Multiply $\frac{1}{x - 2}$ by $\frac{x + 5}{x + 5}$ .

$\frac{x + 5}{(x - 2) (x + 5)} - \frac{2}{x + 5} \cdot \frac{x - 2}{x - 2} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.4.2

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

$\frac{x + 5}{(x - 2) (x + 5)} - \frac{2 (x - 2)}{(x + 5) (x - 2)} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.4.3

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

$\frac{x + 5}{(x + 5) (x - 2)} - \frac{2 (x - 2)}{(x + 5) (x - 2)} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{x + 5 - 2 (x - 2)}{(x + 5) (x - 2)} - \frac{2 x - 1}{(x - 2) (x + 5)} = 0$

Step 2.6

Combine the numerators over the common denominator.

$\frac{x + 5 - 2 (x - 2) - (2 x - 1)}{(x + 5) (x - 2)} = 0$

Step 2.7

Simplify each term.

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Step 2.7.1

Apply the distributive property.

$\frac{x + 5 - 2 x - 2 \cdot - 2 - (2 x - 1)}{(x + 5) (x - 2)} = 0$

Step 2.7.2

Multiply $- 2$ by $- 2$ .

$\frac{x + 5 - 2 x + 4 - (2 x - 1)}{(x + 5) (x - 2)} = 0$

Step 2.7.3

Apply the distributive property.

$\frac{x + 5 - 2 x + 4 - (2 x) - - 1}{(x + 5) (x - 2)} = 0$

Step 2.7.4

Multiply $2$ by $- 1$ .

$\frac{x + 5 - 2 x + 4 - 2 x - - 1}{(x + 5) (x - 2)} = 0$

Step 2.7.5

Multiply $- 1$ by $- 1$ .

$\frac{x + 5 - 2 x + 4 - 2 x + 1}{(x + 5) (x - 2)} = 0$

Step 2.8

Subtract $2 x$ from $x$ .

$\frac{- x + 5 + 4 - 2 x + 1}{(x + 5) (x - 2)} = 0$

Step 2.9

Subtract $2 x$ from $- x$ .

$\frac{- 3 x + 5 + 4 + 1}{(x + 5) (x - 2)} = 0$

Step 2.10

Add $5$ and $4$ .

$\frac{- 3 x + 9 + 1}{(x + 5) (x - 2)} = 0$

Step 2.11

Add $9$ and $1$ .

$\frac{- 3 x + 10}{(x + 5) (x - 2)} = 0$

Step 2.12

Factor $- 1$ out of $- 3 x$ .

$\frac{- (3 x) + 10}{(x + 5) (x - 2)} = 0$

Step 2.13

Rewrite $10$ as $- 1 (- 10)$ .

$\frac{- (3 x) - 1 (- 10)}{(x + 5) (x - 2)} = 0$

Step 2.14

Factor $- 1$ out of $- (3 x) - 1 (- 10)$ .

$\frac{- (3 x - 10)}{(x + 5) (x - 2)} = 0$

Step 2.15

Rewrite $- (3 x - 10)$ as $- 1 (3 x - 10)$ .

$\frac{- 1 (3 x - 10)}{(x + 5) (x - 2)} = 0$

Step 2.16

Move the negative in front of the fraction.

$- \frac{3 x - 10}{(x + 5) (x - 2)} = 0$

Step 3

Set the denominator in $\frac{3 x - 10}{(x + 5) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(x + 5) (x - 2) = 0$

Step 4

Solve for $x$ .

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Step 4.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 5 = 0$

$x - 2 = 0$

Step 4.2

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 4.2.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 4.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 4.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.4

The final solution is all the values that make $(x + 5) (x - 2) = 0$ true.

$x = - 5, 2$

Step 5

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 5, x = 2$

Step 6