# Solving an equation for x

• Jul 26th 2010, 11:58 AM
halfnormalled
Solving an equation for x
Hello folks,

I'm new here (Hello!) so thought I'd better introduce myself. I never took anything further than GSCE level maths at school so I'm trying to teach myself, with the help of some books, calculus and eventually linear algebra. I'm doing pretty good I think and understand 99% of the subjects I come across. But occasionally there are one or two individual examples which just throw me completely and seem to take up a huge amount of my precious study time! I'm sure they are very simple and I'm just missing one tiny obvious thing, but I just have to know how to solve them!

So here goes, my first question: (one of probably many to come)

How can this equation:

$\frac{x+5}{x-1}=e^2$

Be solved for x to give:

$x=\frac{e^2+5}{e^2-1}$

I'm stumped.... (But also hoping you are going to embarrass me with how stupidly easy it is...). Thanks much!

(Hope the latex is ok... my first attempt!)
• Jul 26th 2010, 12:09 PM
Quote:

Originally Posted by halfnormalled
Hello folks,

I'm new here (Hello!) so thought I'd better introduce myself. I never took anything further than GSCE level maths at school so I'm trying to teach myself, with the help of some books, calculus and eventually linear algebra. I'm doing pretty good I think and understand 99% of the subjects I come across. But occasionally there are one or two individual examples which just throw me completely and seem to take up a huge amount of my precious study time! I'm sure they are very simple and I'm just missing one tiny obvious thing, but I just have to know how to solve them!

So here goes, my first question: (one of probably many to come)

How can this equation:

$\frac{x+5}{x-1}=e^2$

Be solved for x to give:

$x=\frac{e^2+5}{e^2-1}$

I'm stumped.... (But also hoping you are going to embarrass me with how stupidly easy it is...). Thanks much!

(Hope the latex is ok... my first attempt!)

You have x in two places but you want it in only one.

If you had x on two sides, you could bring the x's to one side and get x by itself.
Hence the first step is to multiply both sides by the denominator of the fraction...

$\frac{x+5}{x-1}=e^2\ \Rightarrow\ \frac{(x+5)(x-1)}{x-1}=(x-1)e^2$

$x+5=xe^2-e^2$

Get x by itself by subtracting $xe^2$ from both sides

$x-xe^2+5=-e^2$

subtract 5 from both sides

$x-xe^2=-\left(e^2+5\right)$

Now factor the LHS to write a single x
• Jul 27th 2010, 11:00 AM
halfnormalled
Archie, thanks very much! :D It was the factoring out of x which threw me... I keep stumbling on these little things! Just need more practice I guess.