How do I solve simultaneous equations given a linear and a quadratic equation?

Let's use the following example: y - x = 3 y^2 + x^2 = 29 . Firstly, the second equation is the quadratic equation because it has a y-squared term. The first equation is linear because it doesn't have any squared x or y terms. Now, take the linear equation and rewrite it as y = x + 3 (by adding x to both sides of the equation). Then substitute this result into the quadratic equation. What this means is: wherever you see a y in the second equation, instead you write x + 3. This gives: (x + 3)^2 + x^2 = 29 . To simplify this, multiply out the brackets using the following steps: (x + 3)(x + 3) + x^2 = 29, x^2 + 3x + 3x + 9 + x^2 = 29, 2x^2 + 6x - 20 = 0, x^2 + 3x - 10 = 0 . At this point you have a quadratic equation. If you are confident factorising it, this will be the quickest method. Otherwise, you could try using the quadratic formula. You get the following two solutions: x = -5 x = 2 . Substitute these two answers back into the first (the linear) equation. This gives the following solutions: when x = -5, y = -2; when x = 2, y = 5.

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Answered by Erica H. Maths tutor

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