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

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Answer 1 · 2017-03-06T20:01:54

The two solutions are:

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

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

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.

How do you solve y=x^2+7x+12, y=2x+8y=x2+7x+12,y=2x+8y=x^2+7x+12, y=2x+8 using substitution?