```Question 35754
Hi,

We want to find x. At the moment x is in the numerator of one fraction, and the denominator of another. There are a few ways to solve this, but the simplest is to get rid of fractions all together, by multiplying the whole equation by *[tex 10(x-3)]. This gives us

*[tex (x-3)(x+4)=60]

Think about what we've just done if *[tex x=3], Does it matter?

We're now left with this equation, which is your bog standard quadratic. This one can be factorised quite easily. (I miss the days when every quadratic I had to solve was factorisable!)

Multiplying out and tidying up gives

*[tex x^2+x-72=0]

Now, there isn't any coeficient in front of the x, and the constant term is negative, so we're looking at something like:

*[tex (x+?_1)(x-?_2)=0]

Where the question marks multiply to 72, and have a difference of 1. 8, and 9 look like good candidates. So that gives

*[tex (x+9)(x-8)=0]

So from this we can read off that either *[tex x=-9] or *[tex x=8]

Pluging both of these into the original equation shows they both work, so there's your answers.

Hope that helps,
Kev```