Solving systems of equations in 2 variables

OK, so I'm told that to solve this system by substitution, I solve one equation for one
variable and then plug the answer into the other one. Well, here's what I've got so far.

Equation 1: X^2 + Y^2 = 25
Equation 2: 2X + 3Y = -1

So I solve the first equation for X like this?

sqrt X^2 + sqrt Y^2 = sqrt 25 OR X + Y = 5

SO: X = (5 - Y) IS THIS RIGHT?

If so, WHAT NEXT?

Any help is greatly appreciated!

Equation 1: X² + Y² = 25
Equation 2: 2X + 3Y = -1

2X + 3Y = -1
x = - 1/2 - 1 1/2 y

X² + Y² = 25
Y² + X² - 25 = 0
Y² + (- 1/2 - 1 1/2 y)² - 25 = 0
Y² + 1/4 + 1 1/2 y + 2 1/4 y² - 25 = 0
3 1/4 y² + 1 1/2 y - 24 3/4 = 0

3x² + 1x - 24 = 0

ay² + by + c = 0
Discriminant b² - 4ac = 289
y1 = ( -b + square root [ b² - 4ac ] ) : 2a
y1 = 2 2/3
y2 = ( -b - square root [ b² - 4ac ] ) : 2a
y2 = -3

now do X yourself, will ya!

Lets solve equation 2 for x

Equation 2: 2X + 3Y = -1

2X = -1-3Y
X = (-1-3Y)/2

okay we have what x is equal to... now we want to substitute the X in equation 1 with what it is equal to in equation 2.

X^2 + Y^2 = 25
((-1-3Y)/2)^2 + Y^2 = 25

Now that we only have one equation we can solve for Y.

(((-1-3Y)^2)/4)+ Y^2 = 25

FOIL (Firsts, Outers, Inners, Lasts)

(9Y^2 + 6Y +1)/4)+ Y^2 = 25
3.25Y^2 + 1.5Y + .25 - 25 = 0
3.25Y^2 + 1.5Y + .25 - 25 = 0

Now we have a quadratic equation so we can use the quadratic formula for it

Y = -b (+-) (sqrt(b^(2) -4ac) / 2a
Y = (-1.5 (+-) sqrt( (1.5^2) - 4(3.25)(-24.75) )) / 2(3.25)
Y = (-1.5 (+-) 18) / 6.5

Y = 2.5, or Y = 3

X = (-1-3Y)/2
Plug and chug!

X = -4.25, or X = -5