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
Related Maths GCSE answers
All answers ▸Calculate the area of a circle where the circumference of the circle and sides of the square are tangential at 4 points.
b is two thirds of c. 5a = 4c Work out the ratio a : b : c Give your answer in its simplest form where a, b and c are integers
A house increases by 25% to £80,000. Find what it was worth before the rise.
For what values of x is 2x^2 - 11x - 6 > 0 ?