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Step 1: Recognise the quadratic term, this cannot be solved through elimination. Therefore we must need to use substitution. Substitute y = x + 3 into the quadratic. Gives x + 3 = x² + 3xStep 2...

Step 1: Simplify 3n + 2 ≤ 14 3n ≤ 12 n ≤ 4 and 6n > n^2 + 5 0 > n^2 -6...

x+10=0x= -10x-16=0x= 16

Multiplying out the first two brackets gives (x^2-x-12)(x+5). Multiplying the remaining brackets gives x^3 + 4x^2 -17x - 60.

a = 4b = 60

We must use the substitution method for this question because the first equation is a quadratic. Take the more simple equation (the second one), and use that to substitute the value for y in to the first ...