SOLUTION: I am having trouble solving a system of equations: {{{4sqrt(x^2-3x) - 3sqrt(y^2+6y)= -4}}} {{{sqrt(x^2-3x) + sqrt(y^2+6y)= 6}}} I squared both sides to remove the radical:

Algebra ->  Systems-of-equations -> SOLUTION: I am having trouble solving a system of equations: {{{4sqrt(x^2-3x) - 3sqrt(y^2+6y)= -4}}} {{{sqrt(x^2-3x) + sqrt(y^2+6y)= 6}}} I squared both sides to remove the radical:       Log On


   

Question 573236: I am having trouble solving a system of equations:
4sqrt%28x%5E2-3x%29+-+3sqrt%28y%5E2%2B6y%29=+-4
sqrt%28x%5E2-3x%29+%2B+sqrt%28y%5E2%2B6y%29=+6
I squared both sides to remove the radical:
16%28x%5E2-3x%29+-9%28y%5E2%2B6y%29=16
x%5E2-3x%2By%5E2%2B6y=36
and that is as far as I've gotten.
thanks
lpkitty
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
4sqrt%28x%5E2-3x%29+-+3sqrt%28y%5E2%2B6y%29=+-4
sqrt%28x%5E2-3x%29+%2B+sqrt%28y%5E2%2B6y%29=+6
But these are like terms so let's simplify things by using elimination
Multipl the 2nd equation by 3, add to the 1st equation
4sqrt%28x%5E2-3x%29+-+3sqrt%28y%5E2%2B6y%29=+-4
3sqrt%28x%5E2-3x%29+%2B+3sqrt%28y%5E2%2B6y%29=+18
-----------------------------------adding eliminates the y radical, so we have
7sqrt%28x%5E2-3x%29+=+14
divide both sides by 7
sqrt%28x%5E2-3x%29+=+2
square both sides
x^2 - 3x = 4
x^2 - 3x - 4 = 0
Factors to
(x-4)(x+1) = 0
two solutions
x = -1
x = 4
:
Use the 2nd original equation to find y
sqrt%28x%5E2-3x%29+%2B+sqrt%28y%5E2%2B6y%29=+6
sqrt%284%5E2-3%284%29%29+%2B+sqrt%28y%5E2%2B6y%29=+6
sqrt%284%29+%2B+sqrt%28y%5E2%2B6y%29=+6
2+%2B+sqrt%28y%5E2%2B6y%29=+6
sqrt%28y%5E2%2B6y%29=+6-2
sqrt%28y%5E2%2B6y%29=+4
square both sides
y^2 + 6y = 16
y^2 + 6y - 16 = 0
Factors to
(y+8)(y-2) = 0
Y = -8
Y = 2
:
You should check all the solutions in the original equations