Express (4x)/((x^2) - 9) - (2)/(x + 3) as a single fraction in its simplest form (4 marks)

To do this we need to find a common denominator. We have an (x + 3) on the right hand side and a difference of two squares ((x^2)-9) on the left hand side. So factorising what looks more complicated is first priority. The difference of two squares factorises into (x + 3)(x - 3) (this gets the first mark). Now we have (x + 3) in both denominators, but no (x - 3) on the right side. Therefor to get this we must times the right side by (x - 3)/(x - 3) to get (2(x - 3))/(x + 3)(x - 3). We can then combine the two fractions to make one fraction of (4x - 2(x - 3))/(x - 3)(x + 3) (second mark achieved here)
We now have it as a single fraction! However it has asked for our simplest form. First we should expand the numerator as we have an x inside and outside the bracket - this gives us: (4x - 2x + 6)/(x - 3)(x + 3). This then becomes: (2x + 6)/(x - 3)(x + 3) (third mark!). On the numerator we now have a 2x and a 6, these are both multiples of 2, meaning we can take the 2 out, making it: (2(x + 3))/(x + 3)(x - 3). Finally with an (x + 3) on the top and bottom of the fraction we can cancel them out leaving us with our single fraction in its simplest form: 2/(x - 3) (final answer mark!).

LN
Answered by Luke N. Maths tutor

4403 Views

See similar Maths 11 Plus tutors

Related Maths 11 Plus answers

All answers ▸

There are 13 apples in a crate of apples. Mark orders six boxes of apples, how many apples did he order?


The average temperature in May is 18 ºC. The average temperature in November is a third of the average temperature in May. The average temperature in September is twice the average temperature in November. What is the average temperature in September?


I’m confused by the difference between using f’(x) and f”(x) to find the minima and maxima of a curve


The function f is defined by f (x) = e ^(x+2), where x is any real number. Find f ^-1(x) and state its domain.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences