# 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...

-
You lost the most important part somewhere - the equals sign!! Once that is fixed, you will need to "solve a quadratic equation", a skill that you can google/YouTube search. – The Chaz 2.0 Dec 6 '11 at 2:10
There's also at least one multiplication (sign) error. – The Chaz 2.0 Dec 6 '11 at 2:11

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:

$$a\times d=c\times b$$.

-
+1 simple and clear! – Hassan Muhammad Dec 6 '11 at 6:57
typo: (with c,d≠0) should be (with b,d≠0) – Prizoff Jun 2 '14 at 23:56

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

$$\frac{5+8x}{3+2x} = \frac{45-8x}{13-2x} = \frac{50}{16} = \frac{25}{8}$$.