Question

Fan and medal
solve the system of equations
x^2+y^2+2x+2y=0
x^2+y^2+4x+6y+12=0

Answer 1

@AloneS
@Hayhayz

Answer 2

Have you considered simply subtracting the two equations. This gives a single line where the more complicated solutions must lie.

Answer 3

Try to draw these two circles or subtract the first equation from the second as suggested above

Answer 4

Yes, please, complete the square four times and draw the circles. THEN, draw the line given from the subtractions. You will be well on your way.

Answer 5

Is that possible, subtracting the equations?
I know the solutions, I just need to know how to do them

Answer 6

If you subtract them and do some manipulations you get
x=-2y-6
replace in one of the equations and use the quadratic formula

Answer 7

Replace in the first eqaution, you get
\[y^2+2 y+(-2 y-6)^2+2 (-2 y-6)=0
\]

Answer 8

I'm going to guess that you have used the "Elimination Method" many times to solve small linear systems of equations. Why is this system any different? Use the same methods.

x + y = 5
x - 6 = 2

Add them

2x = 7

Done.

Answer 9

Expand the equation above, you get
\[
5 y^2+22 y+24=0
\]

Answer 10

what equation did you expand?

Answer 11

Solving it you get
\[
\left(
\begin{array}{c}
y=-\frac{12}{5} \\
y=-2 \\
\end{array}
\right)
\]

Answer 12

Use x=-2y-6
to find the corresponding x

Answer 13

I expanded the one in my post above

Answer 14

You will get
\[
\left(
\begin{array}{cc}
x=-2 & y=-2 \\
x=-\frac{6}{5} & y=-\frac{12}{5} \\
\end{array}
\right)
\]

Answer 15

You expand
\[
2 (-6 - 2 y) + (-6 - 2 y)^2 + 2 y + y^2
\]

Answer 16

thank you !!!!!

Answer 17

What? We've never heard of an example?