solve for x and y -- x=2y and x^2+2x-y-y^2=36
given x2 + 2x – y – y2 = 36
substitute 2y for every x (2y)2 + (2)(2y) – y – y2 = 36
4y2 + 4y – y – y2 = 36
3y2 + 3y = 36
3y2 + 3y – 36 = 0
(3y – 9)(y + 4) = 0
y = 3 & x = 6
y = –4 & x = –8
check answer
x2 + 2x – y – y2 = 36 x2 + 2x – y – y2 = 36
Does 62 + 12 – 3 – 32 = 36 ? Does (–8)2 +(2)(–8) – (–4) – (–4)2 = 36 ?
36 + 12 – 3 – 9 = 36 64 – 16 + 4 – 16 = 36
36 = 36 True 36 = 36 True
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