# How do you solve \frac { 1} { x - 2} + 3= - \frac { 16} { x ^ { 2} + x - 6}?

Jun 8, 2017

$x = \frac{- 4 \pm \sqrt{388}}{6}$

#### Explanation:

Let's start off with the original equation:

$\frac{1}{x - 2} + 3 = \frac{16}{{x}^{2} + x - 6}$

We then factor ${x}^{2} + x - 6$ into a $\left(x + a\right) \left(x + b\right)$ format:

1/(x-2)+3=16/((x-2)(x+3)

We see that in both fractions, there is an $\left(x - 2\right)$ term, so we multiply by $\left(x - 2\right)$ on both sides:

$\left(x - 2\right) \left(\frac{1}{x - 2} + 3\right) = \left(x - 2\right) \left(\frac{16}{\left(x - 2\right) \left(x + 3\right)}\right)$

Before we can cancel out the $x - 2$ terms however, we need to use the distributive property to expand out the left side:

$\left(x - 2\right) \left(\frac{1}{x - 2}\right) + \left(x - 2\right) \left(3\right) = \left(x - 2\right) \left(\frac{16}{\left(x - 2\right) \left(x + 3\right)}\right)$

Now, we can cancel out the $\left(x - 2\right)$ terms:

$\cancel{x - 2} \left(\frac{1}{\cancel{x - 2}}\right) + \left(x - 2\right) \left(3\right) = \cancel{x - 2} \left(\frac{16}{\cancel{x - 2} \left(x + 3\right)}\right)$

That leaves:

$1 + \left(x - 2\right) \left(3\right) = \frac{16}{x + 3}$

We use the distributive property again on the $\left(x - 2\right) \left(3\right)$ term:

$1 + 3 x - 6 = \frac{16}{x + 3}$

We go ahead and combine the $1$ and the $- 6$ and end up with this:

$3 x - 5 = \frac{16}{x + 3}$

Next, we multiply $\left(x + 3\right)$ on both sides in order to have all the $x$ terms on one side:

$\left(3 x - 5\right) \left(x + 3\right) = 16$

Using FOIL, we can expand out the left side and rewrite the equation as so:

$3 {x}^{2} + 4 x - 15 = 16$

We can then move all the terms to the left side and leave a $0$ on the right:

$3 {x}^{2} + 4 x - 31 = 0$

Here, we can use the Quadratic Formula to obtain the value of $x$. Recalling that the Quadratic Formula is

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

and that the $a$, $b$, and $c$ values in our equation are $3$, $4$, and $- 31$, respectively, we can substitute those values into the formula to find $x$:

$x = - \left(4\right) \pm \frac{\sqrt{{\left(4\right)}^{2} - 4 \left(3\right) \left(- 31\right)}}{2 \left(3\right)}$

$x = - 4 \pm \frac{\sqrt{16 \left(- 12\right) \left(- 31\right)}}{6}$

$x = - 4 \pm \frac{\sqrt{16 + 372}}{6}$

$x = - 4 \pm \frac{\sqrt{388}}{6}$
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$x = - 4 \pm \frac{\sqrt{388}}{6}$ results in two different equations for $x$.

$x = - 4 + \frac{\sqrt{388}}{6}$ and $x = - 4 - \frac{\sqrt{388}}{6}$

This means that $x$ has two possible values:

• $x = - 4 + \frac{\sqrt{388}}{6}$
$x = \frac{- 4 + 19.6977156036}{6}$
$x = \frac{15.6977156036}{6}$
$\textcolor{red}{x = 2.61628593393}$

• $x = - 4 - \frac{\sqrt{388}}{6}$
$x = \frac{- 4 - 19.6977156036}{6}$
$x = \frac{- 23.6977156036}{6}$
$\textcolor{b l u e}{x = - 3.94961926727}$