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### How do I solve simultaneous equations when one of the equations is not linear?

Sometimes you are asked to solve simultaneous equations where one of the equations has an x2 and/or a y2 in it - this is called a non-linear equation. This kind of question can look very complicated but once you learn the techniques it won't seem nearly as hard!

An example of this kind of question would be: Solve the simultaneous equations x2 + y2 = 5 and x + y = 3.

First you would label the first equation (1) and the second equation (2) to make things easier. Then you would make y or x the subject of the linear equation (in this case equation (2)). For example, to make x the subject you would subtract y from both sides to get x = 3 - y. You would then substitute this into equation (1) to get (3 - y)(3 - y) + y2 = 5.

Next you need to solve this equation for y. First you would expand the bracket by multiplying each part of the first bracket by each part of the second bracket to get 9 - 6y + y2 + y2 = 5. Then you would collect like terms to get a quadratic equation in a form you can solve. This would give 2y2 - 6y + 4 = 0. Everything in this equation is divisible by two so you can divide each term by two to give y2 - 3y + 2 = 0. This can be factorised into the form (y + a)(y + b) by looking for two numbers (a and b) that will add to give -3 and multiply to give 2. The only two numbers that will work are -1 and -2 so the equation becomes (y - 1)(y - 2) = 0. Either (y - 1) or (y - 2) must equal 0 and so y is equal to either 1 or 2.

All that's left to do now is substitute the values for y in turn into either equation (1) or equation (2) to find the corresponding values for x. It's normally easier to substitute into the linear equation (i.e. equation (2)) so that is what I'll show here. Taking y = 1, we get x + 1 = 3 and subtracting 1 from each side gives us x = 2. For y = 2, x + 2 = 3 and so subtracting 2 from both sides gives us x = 1.

Therefore the solutions to these simultaneous equations are x = 1, y = 2 and x = 2, y = 1.

2 years ago

Answered by Kristen, who tutored GCSE Maths with MyTutor

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