How do I solve the simultaneous equations x-2y=1 and x^2-xy+y^2=1?

Depending on what you prefer, you can either substitute or you can try to solve this by elimination. I'll explain both. 1) x - 2y = 1 2) x2 - xy +y2 = 1 If solving by elimination, you need to start thinking about how to make the equations similar enough so that at least one of the variables cancels out. Let's try to cancel out x, or x2. Now, if we just multiplied x-2y=1 by x, we would end up nowhere because then we would introduce an x on the right hand side, and wouldn't be able to get rid of the variable. However, by inspection, if we simply squared both sides of the equation, we would get x2-4xy+4y2=1. If we then subtract the second equation from this one, we get 3y2-3xy=0. We can factorise this and write it as 3y(y-x)=0. Now, if the product of two numbers is zero, one of them must be zero. So, either 3y=0 and y=0 or y-x=0 and y=x. If y=0 then x=1 by the first equation. In this case, we can find y or x by substituting y for x in x-2y=1. So y-2y=1 and y=x=-1. The second way is substitution. x-2y=1 can be rewritten as x=1+2y. By substituting this into the second equation, and after doing quite a lot of algebra, we get 1+4y + 4y2-y-2y2+y2=1, or 3y2+3y=0 or 3y(y+1)=0. This means that either y=0 or y=-1. If y=0 we know x=1, and if y=-1 we know x=-1. Use whichever method you prefer.

KS
Answered by Katerina S. Maths tutor

13145 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the simultaneous equations 1) 3x + 2y = 4 & 2) 4x + 5y = 17


Harry drives from Exeter to London in 4 hours at an average speed of 70km/h. Ron drives from Exeter to London in 5 hours. (a) Assuming Ron took the same route as Harry, calculate Ron's average speed.


Insert one pair of brackets so that this calculation is correct; 3 x 6 + 5 - 1 = 32


How to solve a quadratic equation without a calculator?


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:

© 2026 by IXL Learning