Question 573236
{{{4sqrt(x^2-3x) - 3sqrt(y^2+6y)= -4}}}
{{{sqrt(x^2-3x) + sqrt(y^2+6y)= 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(x^2-3x) - 3sqrt(y^2+6y)= -4}}}
{{{3sqrt(x^2-3x) + 3sqrt(y^2+6y)= 18}}}
-----------------------------------adding eliminates the y radical, so we have
{{{7sqrt(x^2-3x) = 14}}}
divide both sides by 7
{{{sqrt(x^2-3x) = 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(x^2-3x) + sqrt(y^2+6y)= 6}}}
{{{sqrt(4^2-3(4)) + sqrt(y^2+6y)= 6}}}
{{{sqrt(4) + sqrt(y^2+6y)= 6}}}
{{{2 + sqrt(y^2+6y)= 6}}}
{{{sqrt(y^2+6y)= 6-2}}}
{{{sqrt(y^2+6y)= 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