Question

$$(\frac{x+y}{x-y}-\frac{x-4}{x+y})\cdot(\frac{x+3}{x-y}+\frac{x-y}{x+y})^{-}$$

Answer

$$-(x*(4-y))/(x*(5-y))^$$

Solution


Collect like terms.
\[((x-x-4x)+yx+(-y+y)){(x+3x-y+x-yx+y)}^{-}\]
Simplify  \((x-x-4x)+yx+(-y+y)\)  to  \(-4x+yx\).
\[(-4x+yx){(x+3x-y+x-yx+y)}^{-}\]
Collect like terms.
\[(-4x+yx){((x+3x+x)+(-y+y)-yx)}^{-}\]
Simplify  \((x+3x+x)+(-y+y)-yx\)  to  \(5x-yx\).
\[(-4x+yx){(5x-yx)}^{-}\]
Factor out the common term \(x\).
\[(-4x+yx){(x(5-y))}^{-}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[(-4x+yx)\times \frac{1}{{(x(5-y))}^{}}\]
Simplify.
\[\frac{-4x+yx}{{(x(5-y))}^{}}\]
Factor out the common term \(x\).
\[\frac{-x(4-y)}{{(x(5-y))}^{}}\]
Move the negative sign to the left.
\[-\frac{x(4-y)}{{(x(5-y))}^{}}\]