Question

How do I solve Y=x^2+ 4x+ 7 and 2x+ Y+ 2=0 ?

Answer 1

You want to know what x equals and what y equals, is it?

Answer 2

yes please

Answer 3

There are two method to solve them, one is elimination, one is substitution. First I will tell you substitution.

Answer 4

ok thank you

Answer 5

It is very easy. The first formula has told you that y=x^2+4x+7. Replace the y in 2x+ Y+ 2=0 with x^2+4x+7, then you get
2x+x^2+4x+7+2=0
x^2+9+6x=0
x^2+3^2+2*3x=0
(x+3)^2=0
\[\sqrt{(x+3)^2}=0\]
\[\pm(x+3)=0\]
If +(x+3)=0, x=0-3, x=-3;
if -(x+3)=0, -x-3=0, -x=3, x=-3 too.
So x=-3.

Now substitute x=-3 into 2x+ Y+ 2=0.
2*(-3)+y+2=0
-6+2=-y
-4=-y
y=4

The solution of the equation is
x=-3
y=4

Answer 6

Can you help me with this question? http://openstudy.com/study#/updates/527e0ae5e4b022369a7c11ea

Answer 7

ill try