Question

Solve for $xandy$

$$\begin{gathered} \frac{2}{x-1}+\frac{y-2}{4}=2; \\ \frac{3}{2(x-1)}+\frac{2(y-2)}{5}=\frac{47}{20} \end{gathered}$$

Answer 1

Step 1:Giving substitution

Given ,

$$\begin{gathered} \frac{2}{x-1}+\frac{y-2}{4}=2 \\ \frac{3}{2(x-1)}+\frac{2(y-2)}{5}=\frac{47}{20} \end{gathered}$$

$$\begin{gathered} Let\frac{1}{x-1}=Aandy-2=B \\ i.e;2A+\frac{B}{4}=2 \\ \Rightarrow 8A+B=8.............................\left(1\right) \\ \frac{3}{2}A+\frac{2}{5}B=\frac{47}{20} \\ LCM=10, \\ \frac{15A+4B}{10}=\frac{47}{20} \\ \Rightarrow 15A+4B=\frac{47}{2} \\ \Rightarrow 30A+8B=47..............................\left(2\right) \end{gathered}$$

Step 2 : Solving for A and B using elimination method

$$\begin{gathered} equation\left(1\right)\times 8\Rightarrow 64A+8B=64..............................\left(3\right) \\ equation\left(2\right)\Rightarrow 30A+8B=47.................................\left(2\right) \end{gathered}$$

$$\begin{gathered} equation\left(3\right)-\left(2\right)\Rightarrow 34A=17 \\ \Rightarrow A=\frac{17}{34}=\frac{1}{2} \\ SubstituteA=\frac{1}{2}in\left(1\right) \\ 1\Rightarrow 8x\frac{1}{2}+B=8 \\ \Rightarrow 4+B=8 \\ \Rightarrow B=8-4=4 \end{gathered}$$

Step 3 : finding x and y

We have ,

$$\begin{gathered} A=\frac{1}{x-1} \\ \Rightarrow \frac{1}{2}=\frac{1}{x-1} \\ \Rightarrow x-1=2 \\ \Rightarrow x=3 \\ B=y-2 \\ \Rightarrow 4=y-2 \\ \Rightarrow y=4+2=6 \end{gathered}$$

Therefore the value of $x=3andy=6$