Solve the equation (2X + 3) / (X-4) - (2X - 8) / (2X + 1) = 1

In order to solve this equation we need to find a common denominator for the 2 fractions. It is impossible to name the common denominator while working with unknown variables such as x,y etc. but we know from working with numbers that in order to find the common denominator of 2 or more fractions we need to find the common multiple of the denominators. In our case the common multiple is simply the product of our two denominators.  Which means ( X - 4 ) x  ( 2X + 1)  is our common denominator. And we need to multiple each fraction by the opposite's denominator. This results in our fraction looking like this     [ ( 2X + 3)( 2X + 1)  - ( 2X - 8)( X - 4) ] / ( X - 4 ) x  ( 2X + 1) 

After multiplying the paranthesis term by term we are left with     (4X+ 8X + 3  -  2X2 + 8X + 8X - 32 ) /  ( X - 4 ) x  ( 2X + 1)   = 1 .  Having a fraction on one side and just a number on the other side means we can simply multiple the number the number with the denominator of the fraction and totally get rid of the denominator. This results in our equation looking like this 2X+ 24X - 29 = 2X2 - 7X -4  moving everything to the left side of the equation and we get 31X -25 = 0 now we can move 25 to the right          31X = 25  and we simply solve for X. In the end X = 25/31  which is roughly 0.81

OG
Answered by Octavian G. Maths tutor

5059 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Find the complex solutions for the following equation: -3x^2+4x+4=0


Find and simplify the point(s) of intersection of the curves: x^2 + y^2 =6 , y = x - 3


A bag has 3 red balls and 5 green balls. I take out 2 balls, without replacing them. What is the probability of choosing at least one red ball? Give your answer to 3 decimal places.


How do you work out the median of a set of numbers?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences