# Solve $\frac{x-a}{b}-\frac{x+c}{d}=0$ for x

I need to solve $$\frac{x-a}{b}-\frac{x+c}{d}=0$$ for x.

$$x=\frac{ad+bc}{d-b}$$

But i can't figure out how to get there, I think i have to start by making the fractions have the same denominator but after that I'm stuck.

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Aurum: You could show the result of making the fractions have the same denominator, this would help the answerers. –  Did Aug 21 '11 at 11:26

You have $$\frac{x}{b} - \frac{a}{b} -\frac{x}{d}-\frac{c}{d} =0$$ From this you get $$\frac{x}{b}-\frac{x}{d} = \frac{a}{b}+\frac{c}{d}$$ $$\Longrightarrow x \Bigl(\frac{d-b}{bd}\Bigr) = \frac{ad+bc}{bd}$$

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Thanks for the help, What i missed when i calculated was to factorize x. And after that it was only to cross multiplicate the fractions and factorize bd and then bd disappears. –  Aurum Aug 21 '11 at 11:43

Bringing the fractions to a common denominator is a very useful idea. Here is a variant of the same idea. $$\frac{x-a}{b}-\frac{x+c}{d}=0$$ precisely if $$bd\left(\frac{x-a}{b}-\frac{x+c}{d}\right)=0.$$ (We are multiplying by $bd$ in order to get rid of the denominators.) But $$bd\left(\frac{x-a}{b}-\frac{x+c}{d}\right)=d(x-a)-b(x+c).$$

So we want to solve the equation $$d(x-a)-b(x+c)=0.$$ Now we are dealing with a much more pleasant expression. Expand. We get $$(d-b)x-ad-bc=0,\quad\text{or equivalently}\quad (d-b)x=ad+bc.$$

If $d \ne b$, we can divide, and obtain $$x=\frac{ad+bc}{d-b}.$$

If $d=b$, we cannot divide. In that case, our equation becomes $(0)x=ad+bc.$ But $d=b$, so we are looking at $(0)x=b(a+c)$. This can only happen if $c=-a$.

So the conclusion is that if $b\ne d$, the solution of the equation is exactly the one you gave. If $b=d$, there is no solution unless $c=-a$. And if $b=d$ and $c=-a$, then every number $x$ is a solution of the equation.

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