# How do you solve (2(x-2))/(x^2-10x+16)=2/(x+2)?

Jun 1, 2017

$x \in \left\{\cancel{\text{O}}\right\}$

(no solutions)

#### Explanation:

First, let's factor ${x}^{2} - 10 x + 16$.

We need two factors of 16 which add up to -10.

$- 2 \mathmr{and} - 8$ work, so our factored polynomial is:

$\left(x - 2\right) \left(x - 8\right)$

Now we can solve the equation:

$\frac{2 \left(x - 2\right)}{{x}^{2} - 10 x + 16} = \frac{2}{x + 2}$

$\frac{2 \left(x - 2\right)}{\left(x - 2\right) \left(x - 8\right)} = \frac{2}{x + 2}$

$\frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 2\right)}}} \left(x - 8\right)} = \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2}}}}{x + 2}$

$\frac{1}{x - 8} = \frac{1}{x + 2}$

Now take the reciprocal of both sides:

$x - 8 = x + 2$

$- 8 = 2$

This is impossible, so there are no solutions.

Jun 1, 2017

No solution

#### Explanation:

$\frac{2 \left(x - 2\right)}{{x}^{2} - 10 x + 16} = \frac{2}{x + 2}$

$\therefore \frac{2 \left({\cancel{x - 2}}^{1}\right)}{\left({\cancel{x - 2}}^{1}\right) \left(x - 8\right)} = \frac{2}{x + 2}$

$\therefore \frac{2}{x - 8} = \frac{2}{x + 2}$

multiply both sides by (x-8)(x+2)

$\therefore 2 \left(x + 2\right) = 2 \left(x - 8\right)$

$\therefore 2 x + 4 = 2 x - 16$

$\therefore 2 x - 2 x = - 16 - 4$

$\therefore 0 = - 20$

No solution