Solve the following set of simultaneous equations: (eq.1) x + 3y = 10, (eq.2) 2x + y = 5

Firstly, multiply eq.1 by 2, to obtain: (eq.3) 2x + 6y = 20Next, subtract eq.2 from eq.3 to obtain: (eq.4) 5y = 15Next, divide eq.4 by 5 to obtain: y = 3Now substitute y = 3 into any of the previous equations, for example, using eq.2 we get: (eq.5) 2x + 3 = 5Now solve eq.5 by...subtracting 3 from both sides: 2x = 2dividing throughout by 2: x = 1Now we have our unique solution to the pair of simultaneous equations: x = 1, y = 3.We can check the solution works by substituting back into one of the first two equations, e.g. in eq.1: x + 3y = 1 + (3x3) = 1 + 9 = 10

JD
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Jo wants to work out the solutions of x^2 + 3x – 5 = 0 She says, ‘‘The solutions cannot be worked out because x^2 + 3x – 5 does not factorise to (x + a)(x + b) where a and b are integers.’’ Is Jo correct?


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