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We have one linear and one quadratic equation here. Since we have a quadratic, we will have two sets of solutions. Let's solve by substitution. Substitute equation (1) into equation (2), which yields:x + 19 = x^2+4x+1OR 0 = x^2 + 3x - 18To solve this quadratic to find our x values we must first factorise, which gives:0 = (x+6)(x-3)It must follow that:x = -6 OR x = 3.Sub these two distinct real solutions into equation 1, we will get our corresponding y value:x = -6, y = 13 OR x=3, y=22.

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One to one online tution can be a great way to brush up on your Maths knowledge.

Have a Free Meeting with one of our hand picked tutors from the UK’s top universities

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