# Maths Exam Questions.

Showing 1 to 6 of 6

Ok for the first question, you are given a quadratic equation over a linear equation and I am assuming you want to rearragne the equation to find it in its simplest form. The way i would look at this is by trying to get an (x+2) term on the top because (x+2)/(x+2) =1 so you would get 1 times the remaining term on the top.

First of all then, you want to factorise the x squared term on the top.

I normally do it like this (x + ) ( x + )

and try to think what the two empty numbers could be. We know these numbers have to multiply together to get 6. and also must be x(a+b)=5x

so the two numbers added together must equal 5.

This means it has to be 2 and 3 so our numerator can look like this (x+2)(x+3). and we now have the equation in the form that we want it in with an (x+2) on the top.

(x+2)(x+3)/(x+2) = 1 (x+3)= (x+3)

So simply factorise the top term so you can divide it by the bottom term, If its a 'top heavy' fraction. In GCSE you will not be asked to use long division i believe so its safe to assume all questions will be of a similar style to this. hope this helps.

Second question is the same style to the first question, we want to simplify the fraction by trying to get the same thing on the bottom as the top.

factorising 4a-20

take out a 4 and we get 4(a-5)

as 4 multiplied by a is 4a and -5 multiplied by 4 = -20

factorising a^2 - 25

squared term so (a+ ) (a+ )

what numbers times together makes -25? and a( b+ c) = 0 the only numbers which satisfy this is 5 and -5.

So (a+5)(a-5)

therefore we get 4(a-5)/(a+5)(a-5).

(a-5) cancels so we are left with 4/(a+5) which is the fraction in its simplest form.

0.29292929292929292 = 1x 0.292929292929

Then let 29.2929292929292929 = 100x 0.292929292929292929

Then if we take away 0.292929 from 29.2929292 then we are left with 29 or (100-1)x 0.29292929292929

so 29 = 99x (0.29292929) so 0.292929292929 as a fraction is 29/99