How do you solve #y=x^2+7x+12, y=2x+8# using substitution?
1 Answer
The two solutions are:
#(x, y) = (-4, 0)#
#(x, y) = (-1, 6)#
Explanation:
Given:
#{ (y = x^2+7x+12), (y = 2x+8) :}#
From the second equation we know that
#2x+8 = x^2+7x+12#
Subtract
#0 = x^2+5x+4#
#color(white)(0) = (x+4)(x+1)#
So
If
If
So the two solutions are:
#(x, y) = (-4, 0)#
#(x, y) = (-1, 6)#
Footnote
Note that it would have been a tiny bit quicker to simply subtract the second given equation from the first to get:
#0 = x^2+5x+4#
#color(white)(0) = (x+4)(x+1)#
Substitution is more useful when one of the given equations uses the substituted variable in a more complicated way.