Question

Solve for $xandy$

$\frac{x-4}{x-3}=\frac{y+4}{y+7};\frac{x+5}{x+2}=\frac{y-1}{y-2}$

Answer 1

Step 1 : Solving using cross multiplication method

Given,

$$\begin{gathered} \frac{x-4}{x-3}=\frac{y+4}{y+7}..............................................\left(1\right) \\ \frac{x+5}{x+2}=\frac{y-1}{y-2}..............................................\left(2\right) \end{gathered}$$

We have,

$$\begin{gathered} \frac{x-4}{x-3}=\frac{y+4}{y+7} \\ \Rightarrow (x-4)(y+7)=(x-3)(y+4) \\ \Rightarrow xy+7x-4y-28=xy+4x-3y-12 \\ \Rightarrow xy+7x-4y-28-xy-4x+3y+12=0 \\ \Rightarrow 3x-y-16=0........................................\left(3\right) \\ Wehave, \\ \frac{x+5}{x+2}=\frac{y-1}{y-2} \\ \Rightarrow (x+5)(y-2)=(x+2)(y-1) \\ \Rightarrow xy-2x+5y-10=xy-x+2y-2 \\ \Rightarrow xy-2x+5y-10-xy+x-2y+2=0 \\ \Rightarrow -x+3y-8=0.................................\left(4\right) \end{gathered}$$

Step 2 : Solving for x and y

$$\begin{gathered} equation\left(3\right)\Rightarrow 3x-y=16 \\ equation\left(4\right)\Rightarrow -x+3y=8 \\ \left(3\right)\times 3\Rightarrow 9x-3y=48.................................\left(5\right) \\ \left(4\right)\Rightarrow -x+3y=8......................................\left(4\right) \\ \left(5\right)+\left(4\right)\Rightarrow 8x=56 \\ \Rightarrow x=\frac{56}{8}=7 \\ equation\left(4\right)\Rightarrow 3y=8+x \\ =8+7=15 \\ \Rightarrow y=5 \end{gathered}$$

Therefore the values of $x=7andy=5$