Question

Please help me with this easy question. How do I find x and y from these 2 equations:

3x^2 + 3y^2 - 15 = 0
6xy + 3y^2 - 15 = 0

I'm having a brain fart right now.

Answer 1

equate them

Answer 2

3x^2 + 3y^2 - 15 = 6xy + 3y^2 - 15

3x^2 + 3y^2 = 6xy + 3y^2

3x^2 = 6xy

3x^2 - 6xy = 0

3x(x - 2y) = 0

when x = 0 or y = x/2

Answer 3

3x^2 +0xy + 3y^2 - 15
0x^2 +6xy + 3y^2 - 15


1x^2 +0xy + 1y^2 - 5
0x^2 +6xy + 3y^2 - 15


1x^2 +0xy + 1y^2 - 5
0x^2 +1xy + 1/2y^2 - 5/2

x^2 = 5 - y^2
xy = (5-y^2)/2

Answer 4

subbing in for x = sqrt(5-y^2)

(24^2+1)y^2 - 10y + (25-5(24^2)) = 0

y = (10 +- sqrt(100-4(24^2+1)(25-5(24^2))))/(2(24^2+1))

y = abt 2.233 or -2.216
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y^2 = 5 - x^2

xsqrt(5-x^2) = -x^2/2
x^2(5-x^2) = -x^4/4
4x^2(5-x^2) = -x^4

20x^2 -4x^4 = -x^4

20x^2(20 -3x^2) = 0

x = 0, +-sqrt(20/3)

maybe

Answer 5

hmm, the solution says the points should be (2,1) , (-2,1) , (0, sqrt(5)) I think it should be something more obvious but I'm missing it..

Answer 6

when x=0 is a fer sure

3(0)^2 + 3y^2 - 15 = 0
6(0)y + 3y^2 - 15 = 0

3y^2 = 15; y = +- sqrt(5)

Answer 7

3x^2 + 3y^2 - 15 = 0

6xy + 3y^2 - 15 = 0
x = (15-3y^2)/6y might be a good sub

Answer 8

so solve for x on the 2nd equaton and plug into 1st?

Answer 9

yeah

3((15-3y^2)/6y)^2 + 3y^2 = 15

(15-3y^2)/6y)^2 + y^2 = 5

(15^2+9y^4-45y^2)/36y^2 + y^2 = 5

15^2+9y^4-45y^2 + 36y^4 = 5(36)y^2

5 - y^2 + y^4 = 4y^2

5 - 3y^2 + y^4 = 0

Answer 10

solving the qudratic gets us a set: {-sqrt(5),-2,-1,1,2} to test out

Answer 11

Ok thank you so much I will rework it

Answer 12

good luck