How do you solve #y=x^2+7x+12, y=2x+8# using substitution?

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Answer 1 · 2017-03-06T20:01:54

The two solutions are:

#(x, y) = (-4, 0)#

#(x, y) = (-1, 6)#

Given:

#{ (y = x^2+7x+12), (y = 2x+8) :}#

From the second equation we know that #y = 2x+8#, so we can substitute #2x+8# for #y# in the first equation to get:

#2x+8 = x^2+7x+12#

Subtract #2x+8# from both sides to get:

#0 = x^2+5x+4#

#color(white)(0) = (x+4)(x+1)#

So #x = -4" "# or #" "x = -1#

If #x=-4# then #y = 2x+8 = 2(-4)+8 = 0#

If #x=-1# then #y = 2x+8 = 2(-1)+8 = 6#

So the two solutions are:

#(x, y) = (-4, 0)#

#(x, y) = (-1, 6)#

#color(white)()#
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.