# How do you solve (x+2)/(x+3)-1=1/(3-2x-x^2) and check for extraneous solutions?

May 13, 2017

#### Explanation:

Given: $\frac{x + 2}{x + 3} - 1 = \frac{1}{3 - 2 x - {x}^{2}}$

Multiply the right side by 1 in the form of $\frac{- 1}{- 1}$:

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

Factor the right side's denominator:

$\frac{x + 2}{x + 3} - 1 = - \frac{1}{\left(x - 1\right) \left(x + 3\right)}$

Restrict the domain to prevent division by 0:

(x+2)/(x+3)-1=-1/((x-1)(x+3)); x!=1, x!=-3

NOTE: We restrict the values of x to make us notice, if we obtain an extraneous root.

Multiply both sides by $x + 3$

(x+2)-x - 3=-1/(x-1); x!=1, x!=-3

Combine like terms:

-1 = -1/(x-1); x!=1, x!=-3

Multiply both sides by $- \left(x - 1\right)$:

x-1 = 1; x!=1, x!=-3

Add 1 to both sides and drop the restrictions, because the solution will not violate them.

$x = 2$

Check:

$\frac{2 + 2}{2 + 3} - 1 = \frac{1}{3 - 2 \left(2\right) - {2}^{2}}$

$\frac{4}{5} - 1 = \frac{1}{-} 5$

$- \frac{1}{5} = - \frac{1}{5}$

This checks.