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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.

  • 2
    The$r$e's also at least o$n$e multiplication (sign) error.2011-12-06

3 Answers 3

1

Move everything to one side of the equality sign. Then you will get a linear equation.

6

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 :).

2

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.