SOLUTION: solve algebraically 4x^2+y^2=64 x^2+y^2=52

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Question 931592: solve algebraically
4x^2+y^2=64
x^2+y^2=52
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2+y^2=64
-1(x^2+y^2=52)
3x^2 = 12
x^2 = 4
x = ± 2
y = √48 = ±4√3 y%5E2+=+52-4
.......
(-2,4√3), (-2,-4√3), (2,4√3), (2,-4√3),


Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
4x%5E2%2By%5E2=64....eq.1
x%5E2%2By%5E2=52.....eq.2
___________________________subtract eq.2 from eq.1
4x%5E2%2By%5E2-%28x%5E2%2By%5E2%29=64-52
4x%5E2%2Bcross%28y%5E2%29-x%5E2-cross%28y%5E2%29=12
3x%5E2=12
x%5E2=12%2F3
x%5E2=4
x=sqrt%284%29
solutions: x=2 or x=-2
substitute these solutions in eq.1 or eq.2 and solve for y
2%5E2%2By%5E2=52.....eq.2
4%2By%5E2=52
y%5E2=52-4
y%5E2=48
y=sqrt%2848%29
solutions: y=6.93 or y=-6.93
system solutions:
x=2 and y=6.93
x=2 and y=-6.93
or
x=-2 and y=6.93
x=-2 and y=-6.93