Question 1034957
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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)    {{{y^2 + 2y - 8}}} = {{{0}}}.   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^3 + x^2}}} = {{{0}}}.   
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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