Trying to solve this simple algebra problems: $\frac{5 + 8x}{3 + 2x} = \frac{45 - 8x}{13 - 2x}$ I know it's kind of stupid to ask this question. But I have problems to solve this simple problems. Can someone point me to the right direction? Did I do something wrong in the process or it's a stupid mistake.... :-( 
Here's the problem:
$$ \frac{5 + 8x}{3 + 2x}  = \frac{45 - 8x}{13 - 2x} $$
I know that I have to cross multiply
$$ \frac{(5 + 8x) \cdot (13 - 2x)}{(3 + 2x) \cdot (45 - 8x)} $$
$$ \frac{65 + 114x - 16x^2}{135 + 66x - 16x^2} $$
Now I do not know how to solve it...
Thank you in advance.
 A: A neat trick for these: 
If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a}{b} = \frac{c}{d} = \frac{a+c}{b+d}$, so 
$$\frac{5+8x}{3+2x} = \frac{45-8x}{13-2x} = \frac{50}{16} = \frac{25}{8}$$.
Now you don't have to worry about quadratic terms :).  
A: Your problem:
First considerations:
$ \frac{5+8x}{3+2x}=\frac{45−8x}{13−2x}$
First of all you need check the existence conditions of the denominator, because, it can't be $0$.
So,
$3+2x\neq0 \implies 2x\neq-3 \implies x\neq\frac{-3}{2}$
and
$13-2x\neq0 \implies -2x\neq-13 \implies x\neq\frac{-13}{-2} \implies x\neq\frac{13}{2}$
This shows the $x=\frac{-3}{2}$ and $x=\frac{13}{2}$ can't be solutions for the equation.
Now is the hint:
If you have two fractions like:
$$\frac{a}{b}=\frac{c}{d} \quad (\text{with} \quad b,d\neq0)$$
You can first multiply the first denominator ($b$) in both sides:
$$a=\frac{c\times b}{d}$$
And finally multiply the second denominator ($d$) in both sides, too:
$$a\times d=c\times b$$.
I think with this you can go ahead^^. If you have difficult yet, post a comment.
A: Move everything to one side of the equality sign. Then you will get a linear equation.
