Solve by Factoring x/(x-2)+2/(x+3)=10/(x^2+x-6)

$\frac{x}{x - 2} + \frac{2}{x + 3} = \frac{10}{x^{2} + x - 6}$

Step 1

Subtract $\frac{10}{x^{2} + x - 6}$ from both sides of the equation.

$\frac{x}{x - 2} + \frac{2}{x + 3} - \frac{10}{x^{2} + x - 6} = 0$

Step 2

Simplify $\frac{x}{x - 2} + \frac{2}{x + 3} - \frac{10}{x^{2} + x - 6}$ .

Factor $x^{2} + x - 6$ using the AC method.

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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

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

To write $\frac{x}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

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

To write $\frac{2}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

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

Write each expression with a common denominator of $(x - 2) (x + 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{x - 2}$ by $\frac{x + 3}{x + 3}$ .

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

Multiply $\frac{2}{x + 3}$ by $\frac{x - 2}{x - 2}$ .

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

Reorder the factors of $(x - 2) (x + 3)$ .

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

Combine the numerators over the common denominator.

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $x$ by $x$ .

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

Move $3$ to the left of $x$ .

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

Apply the distributive property.

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

Multiply $2$ by $- 2$ .

$\frac{x^{2} + 3 x + 2 x - 4 - 10}{(x + 3) (x - 2)} = 0$

Add $3 x$ and $2 x$ .

$\frac{x^{2} + 5 x - 4 - 10}{(x + 3) (x - 2)} = 0$

Subtract $10$ from $- 4$ .

$\frac{x^{2} + 5 x - 14}{(x + 3) (x - 2)} = 0$

Factor $x^{2} + 5 x - 14$ using the AC method.

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 $- 14$ and whose sum is $5$ .

$- 2, 7$

Write the factored form using these integers.

$\frac{(x - 2) (x + 7)}{(x + 3) (x - 2)} = 0$

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{(x - 2) (x + 7)}{(x + 3) (x - 2)} = 0$

Rewrite the expression.

$\frac{x + 7}{x + 3} = 0$

Step 3

Set the numerator equal to zero.

$x + 7 = 0$

Step 4

Subtract $7$ from both sides of the equation.

$x = - 7$

Solve by Factoring x/(x-2)+2/(x+3)=10/(x^2+x-6)

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