SOLUTION: If anyone could please help me to solve these four problems it would be awesome! I am extremely lost. Please help! Thanks!! They use the Substituion Method & Addition Method. I can

Question 430710: If anyone could please help me to solve these four problems it would be awesome! I am extremely lost. Please help! Thanks!! They use the Substituion Method & Addition Method. I can't figure out the solution sets. Thank you!
#1 Substituion Method:
x+y=1
x^2+xy-y^2=-11
#2 Substituion Method:
x^2+y^2=180
x-y=-6
#3 Addition Method:
2x^2+3y^2-30=0
5x^2-7y^2-17=0
#4 Addition Method:
x^2+y^2=16
y^2-3x=16
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
#1 Substituion Method:
x+y=1
x^2+xy-y^2=-11
---
y = 1-x
Substitute:
x^2+x(1-x)-(1-x)^2 = -11
----
x^2+x-x^2-(1-2x+x^2) = -11
---
x-1+2x-x^2 = -11
---
x^2-3x-10 = 0
(x-5)(x+2) = 0
x = 5 or x = -2
If x = 5, y = -4
If x = -2, y = 3
==========================

#2 Substituion Method:
x^2+y^2=180
x-y=-6
---
x = y-6
Substitute:
(y-6)^2+y^2 = 180
y^2-12y+36+y^2 = 180
---
2y^2-12y-144 = 0
y^2-6y-72 = 0
(y-12)(y+6) = 0
y = 12 or y = -6
----
If y = 12, x = 6
If y = -6, y -12
=========================
#3 Addition Method:
2x^2+3y^2-30=0
5x^2-7y^2-17=0
---
Multiply thru the top equation by 5
Multiply thru the bottom equation by 2
---
Subtract, then solve for "y".
Then solve for "x".
-------------------------
#4 Addition Method:
x^2+y^2=16
y^2-3x=16
---
Subtract to get:
x^2+3x=0
x(x+3) = 0
x = 0 or x = -3
---
If x = 0, y = 4 or -4
If x = -3, y = sqrt(7) or -sqrt(7)
=========================================
Cheers,
Stan H.

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