Solve the following simultaneous equations: 2a-5b=11, 3a+2b=7

Let 2a-5b=11 be Equation 1 and 3a+2b=7 be Equation 2. To find a and b, we first need to eliminate one of these variables from the equation. Firstly we can eliminate a from both equations to find b. To do this, we can multiply Equation 1 by 3 and Equation 2 by 2. This gives us: 6a-15b=33, 6a+4b=14. If we take away Equation 1 from Equation 2, we are left with: -15b-4b=33-14. Solving this gives: -19b=19, b=-1. Now that we have obtained b, we can substitute this value back into one of our original equations to obtain a: 2a-5b=11, 2a+5=11, 2a=6, a=3. Hence a=3, b=-1. Note:You can also solve these equations by elimination b first rather than a, you will still obtain the same answer.

GC
Answered by Gemma C. Maths tutor

15728 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How would you differentiate a function such as 4x^3


Solve the simultaneous equations '2X+Y=7' and '3X-Y=8'


A pyramid has a square base with sides of length 4m and a height 3m. What is the length from one of the base corners to the top of the pyramid?


y = 4x^2 + 20x + 11 is a curve. Find the minimum point of the curve.


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning