Question

Solve for $xandy$

$\frac{x+y+3}{x-y-3}=\frac{-3}{2};\frac{x-y-3}{x-y+3}=-2$

Answer 1

Step 1 : Simplifying the equation by cross multiplication method

$$\begin{gathered} \frac{x+y+3}{x-y-3}=\frac{-3}{2} \\ \Rightarrow 2(x+y+3)=-3(x-y-3) \\ \Rightarrow 2x+2y+6=-3x+3y+9 \\ \Rightarrow 2x+2y+6+3x-3y-9=0 \\ \Rightarrow 5x-y-3=0 \\ \Rightarrow 5x-y=3..................................\left(1\right) \\ Also,\frac{x-y-3}{x-y+3}=-2 \\ \Rightarrow x-y-3=-2(x-y+3) \\ \Rightarrow x-y-3=-2x+2y-6 \\ \Rightarrow x-y-3+2x-2y+6=0 \\ \Rightarrow 3x-3y+3=0 \\ \Rightarrow x-y=-1...............................\left(2\right) \end{gathered}$$

Step 2 : Solving (1) and (2) by using elimination method

$$\begin{gathered} Equation\left(1\right)-\left(2\right)\Rightarrow 4x=4 \\ \Rightarrow x=1 \\ substitutex=1inequation\left(2\right) \\ Equation\left(2\right)\Rightarrow 1-y=-1 \\ \Rightarrow -y=-2 \\ \Rightarrow y=2 \end{gathered}$$

Therefore the values of $x=1andy=2$