SOLUTION: Help with pre cal substitution questions!! Solve Using Substitution 2x+4=4 (X+1)^2 + (y-2)^2 = 4 Solve using addition x^2-4y^2 = -7 3x^2 + y^2 =31 x^2 - 2y = 8

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Question 1034957: Help with pre cal substitution questions!!
Solve Using Substitution
2x+4=4
(X+1)^2 + (y-2)^2 = 4
Solve using addition
x^2-4y^2 = -7
3x^2 + y^2 =31

x^2 - 2y = 8
x^2 + y^2 = 16

Solve each system by the method of your choice
2x^2 + y^2 = 18
xy=4


x^3 + y = 0
x^2 - y = 0

x^3 + y = 0
2x^2 - y = 0


x^2 - y^2 - 4x + 6y - 4 = 0
x^2 + y^2 - 4x - 6y + 12 = 0

Answer by ikleyn(52802)   (Show Source): You can put this solution on YOUR website!
.
Solve Using Substitution 

2x+4=4
(X+1)^2 + (y-2)^2 = 4


Solve using addition 

x^2-4y^2 = -7
3x^2 + y^2 =31



x^2 - 2y = 8       (1)    Distract (1) from (2). You will get
x^2 + y^2 = 16     (2)     = .   Solve it using quadratic formula  or  factor.



Solve each system by the method of your choice 
2x^2 + y^2 = 18
xy=4



x^3 + y = 0      (1)     Add (1) and (2). You will get  = .   
x^2 - y = 0      (2)     Solve it by factoring.



x^3 + y = 0      (1)     Same as above
2x^2 - y = 0     (2) 



x^2 - y^2 - 4x + 6y - 4 = 0     (1)     Add (1) and (2) 
x^2 + y^2 - 4x - 6y + 12 = 0    (2)


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