# How do you solve \frac { 4} { x - 3} + \frac { 3} { x + 9} = \frac { 6} { ( x + 9) ( x - 3) }?

Jun 13, 2017

$x = - 3$

But, $x$ cannot equal to $3$ nor $- 9$.

#### Explanation:

$\frac{4}{x - 3} + \frac{3}{x + 9} = \frac{6}{\left(x + 9\right) \left(x - 3\right)}$

Seeing as how there are fractions, we can multiply by both sides $\left(x - 3\right) \left(x + 9\right)$. This has the purpose of removing the $\left(x - 3\right)$ and the $\left(x + 9\right)$ on the left side and the $\left(x - 3\right) \left(x + 9\right)$ on the right side:

$\left(x - 3\right) \left(x + 9\right) \left(\frac{4}{x - 3} + \frac{3}{x + 9}\right) = \left(x - 3\right) \left(x + 9\right) \left(\frac{6}{\left(x + 9\right) \left(x - 3\right)}\right)$

Then we can use the Distributive Property to simplify out the left side:

$\left(x - 3\right) \left(x + 9\right) \left(\frac{4}{x - 3}\right) + \left(x - 3\right) \left(x + 9\right) \left(\frac{3}{x + 9}\right) = \left(x - 3\right) \left(x + 9\right) \left(\frac{6}{\left(x + 9\right) \left(x - 3\right)}\right)$

Now we can cancel out some terms and then simplify the result:

$\cancel{x - 3} \left(x + 9\right) \left(\frac{4}{\cancel{x - 3}}\right) + \left(x - 3\right) \cancel{x + 9} \left(\frac{3}{\cancel{x + 9}}\right) = \cancel{\left(x - 3\right) \left(x + 9\right)} \left(\frac{6}{\cancel{\left(x + 9\right) \left(x - 3\right)}}\right)$

$4 \left(x + 9\right) + 3 \left(x - 3\right) = 6$

We can use the Distributive Property again and simplify the equation:

$4 x + 36 + 3 x - 9 = 6$

We then combine like terms and move all the $x$ terms to the left and the numerical terms to the right:

$7 x + 27 = 6$

$7 x = - 21$

We then divide both sides by $7$ to isolate $x$:

$x = - 3$

However, we're not done. We have to make sure that the value of our variable won't make any of the denominators in the original equation turn to $0$ because otherwise, the solution will be an extraneous solution.

So, when $x = - 3$,

• $x - 3 = - 3 - 3 = \textcolor{red}{- 6}$
• $x + 9 = - 3 + 9 = \textcolor{red}{6}$
• $\left(x + 9\right) \left(x - 3\right) = \left(- 3 + 9\right) \left(- 3 - 3\right) = \left(6\right) \left(- 6\right) = \textcolor{red}{- 36}$

None of these denominators are equal to $0$ so therefore, our solution is $x = - 3$.

However, there are some solutions that do make the denominators $0$. For example:

• For the denominator $x - 3$, if $\textcolor{b l u e}{x = 3}$, then the denominator will turn to $0$, making $3$ a value that $x$ cannot be.

• For the denominator $x + 9$, if $\textcolor{b l u e}{x = - 9}$, then the denominator will turn to $0$, making $- 9$ a value that $x$ cannot be.

• For the denominator $\left(x - 3\right) \left(x + 9\right)$, if $\textcolor{b l u e}{x = 3}$ or $\textcolor{b l u e}{x = - 9}$, then the denominator will turn to $0$, making $3$ and $- 9$ values that $x$ cannot be.

In conclusion, $x$ cannot be $3$ nor $- 9$, making these two values values $x$ cannot be.

Hope this helped!