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Solve the following simultaneous equations?
y=x^2+3x+2
y=2x+8
and
y-x^2-3x+2
y=x+1
With steps please
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- 8 years agoFavorite Answer
y=x^2+3x+2 and y=2x+8
therefore x^2=3x+2 = 2x+8
x^2+3x+2-2x-8 = 0
x^2+x-6 = 0
x^2+3x-2x-6 = 0
x(x+3)-2(x+3) = 0
(x+3)(x-2) = 0
x+3 = 0 or x-2 = 0
x = -3 or x = 2
when x = -3, y = 2(-3)+8 = -6+8 = 2
when x= 2, y = 2(2)+8 = 4+8 = 12
solutions are {-3,2} and {2, 12}
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