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.

Answered by Katerina S. Maths tutor

10904 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Write 120 as a product of its prime factor


Factorise 3xy+6x^2


Factorise h^2 - 36.


If Tom flips a coin 3 times, what is the probability the result are heads, heads, heads, or tails,heads, tails? (assuming we have a fair coin)


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 2013–2024

Terms & Conditions|Privacy Policy