Question

Solve for x: 2x-3y%7D%7Bx%5E2-y%5E2%7D-2x%2B3y%29" id="TexFormula1" title="\frac{3}{x+y} -\frac{3x-3y}{2x-3y} *(\frac{2x-3y}{x^2-y^2}-2x+3y)" alt="\frac{3}{x+y} -\frac{3x-3y}{2x-3y} *(\frac{2x-3y}{x^2-y^2}-2x+3y)" align="absmiddle" class="latex-formula">

Answer 1

Answer:

x = -(3 y)/2

Step-by-step explanation:

Solve for x:

-2 x - 3 y + 3/(x + y) - (3 x - 3 y)/(x^2 - y^2) = 0

Simplify and substitute z = x + y.

-2 x - 3 y + 3/(x + y) - (3 x - 3 y)/(x^2 - y^2) = -y - 2 (x + y)

= -y - 2 z:

-y - 2 z = 0

Multiply both sides by -1:

y + 2 z = 0

Subtract y from both sides:

2 z = -y

Divide both sides by 2:

z = -y/2

Substitute back for z = x + y:

x + y = -y/2

Subtract y from both sides:

Answer:  x = -(3 y)/2