# Thread: big test could do with some help

1. ## big test could do with some help

i started a A level maths course and have a big half way test and would like some pointers in the right direction please thank you for any help
(x-y)( 1 + 1 )
x y

on this question do i find a common donominator so i get y over xy +x over xy so i would have (x-y)(y+x) then multiply out from here ?

next question
(2p-3q)^2 + (2p-3q)
so i get to the point
4p^2-6pq-6qp+9q^2 +(2p-3q) is this it or what else do i do any help please

and last question i dont even know how to start this

m+1 - m-2
2m-3 2m+3

thank you 4 any help

2. Hi

1) $(x - y)(\frac{1}{x} + \frac{1}{y}) = (x - y)(\frac{y}{xy} + \frac{x}{xy}) = (x - y)\,\frac{x + y}{xy}$

And (x-y)(x+y)=x²-y²

2) it depends what the question is !
If you need to develop then it is OK
If you need to factor then
$(2p-3q)^2 + (2p-3q) = (2p-3q)(2p-3q+1)$

3) $\frac{m+1}{2m-3} - \frac{m-2}{2m+3} = \frac{(m+1)(2m+3)}{(2m-3)(2m+3)} - \frac{(m-2)(2m-3)}{(2m+3)(2m-3)}$

$\frac{m+1}{2m-3} - \frac{m-2}{2m+3} = \frac{(m+1)(2m+3)-(m-2)(2m-3)}{(2m-3)(2m+3)}$

$\frac{m+1}{2m-3} - \frac{m-2}{2m+3} = \frac{(2m^2+5m+3)-(2m^2-7m+6)}{(2m-3)(2m+3)}$

$\frac{m+1}{2m-3} - \frac{m-2}{2m+3} = \frac{12m-3}{(2m-3)(2m+3)} = 3 \, \frac{4m-1}{(2m-3)(2m+3)}$