solve the system of equations:?

x^2 - y = 0 (1)
2x + y = 0 (2)
----------
solve by substitution...
x^2 = y (1)
2x + y = 0 (2)

substitute into the second equation the value of y in 1
2x + x^2 = 0

factor out the x
x(2+x) = 0

you get 2 parts...
x = 0
2+x = 0
so... x = 0, -2

plug it into either equation to solve for the y (either equation works)..
2(0) + y = 0
y = 0
(0,0)

2(-2) + y = 0
y = 4
(-2,4)

=============
answers: (0,0), (-2,4)

i know the answer to all math equation


E=mc2

x^2 - y = 0 (solve by using substitution)
2x + y = 0

x^2 - y = 0
y = x^2

2x + y = 0
2x + x^2 = 0
x(2 + x) = 0

x = 0

2 + x = 0
x = -2

x^2 - y = 0
0^2 - y = 0
0 - y = 0
-y = 0
y = 0/-1
y = 0

x^2 - y = 0
-2^2 - y = 0
-y = -4
y = -4/-1
y = 4

∴ (x = 0 , y = 0) , (x = 2 , y = -4)

x² + 2x = 0
x (x + 2) = 0
x = 0 , x = - 2
y = 0 , y = 4

(0,0) , (- 2,4)

x^2 - y = 0
+ 2x + y = 0
x^2 + 2x = 0
x ( x + 2 ) = 0
x = 0 ; x = -2
substitute the values of x...
if x = 0;
2 ( 0 ) + y = 0
y = 0
if x = -2;
2 ( -2 ) + y = 0
-4 + y = 0
y = 4
so we have the solutionset which is { (0,0) , (-2,4) }

x(-2)^2=4
4-y(4)=0
2x(-2)=4

y=4 x=-2

y = x^2

y = -2x

Therefore: x^2 = -2x

=> x(x +2) = 0

x= 0 or x = -2
y= 0 or y = 4

I'll represent x^2 as X
X-y=0
then X=y
then 2x+X=0
then x(x+2)=0
then x=0 or x+2=0
then x=2 or x= -2

x^2-y=0
x^2=y
2x+y=0
2x=-y
x=-y/2