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

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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(52797) (Show Source):
You can put this solution on YOUR website!
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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)

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