Question

Solve for $xandy$

$\frac{16}{x+3}+\frac{3}{y-2}=5;\frac{8}{x+3}-\frac{1}{y-2}=0$

Answer 1

Step 1 : Giving substitution

Given,

$$\begin{gathered} \frac{16}{x+3}+\frac{3}{y-2}=5........................................\left(1\right) \\ \frac{8}{x+3}-\frac{1}{y-2}=0...........................................\left(2\right) \end{gathered}$$

$$\begin{gathered} let\frac{1}{x+3}=A\&\frac{1}{y-2}=B \\ 1\Rightarrow 16A+3B=5.........................................\left(3\right) \\ 2\Rightarrow 8A-B=0.................................................\left(4\right) \end{gathered}$$

Step 2 : To find A & B using elimination method

$$\begin{gathered} \left(4\right)x\left(3\right)\Rightarrow 24A-3B=0........................................\left(5\right) \\ \left(3\right)\Rightarrow 16A+3B=5...............................................\left(3\right) \\ \left(5\right)+\left(3\right)\Rightarrow 40A=5 \\ \Rightarrow A=\frac{5}{40}=\frac{1}{8} \\ SubstituteA=\frac{1}{8}in\left(4\right) \\ \left(4\right)\Rightarrow 8x\frac{1}{8}-B=0 \\ \Rightarrow 1-B=0 \\ \Rightarrow B=1 \end{gathered}$$

Step 3: To find x and y

We have

$$\begin{gathered} WehaveA=\frac{1}{x+3} \\ \Rightarrow \frac{1}{8}=\frac{1}{X+3} \\ \Rightarrow x+3=8 \\ \Rightarrow x=5 \\ B=\frac{1}{y-2} \\ \Rightarrow 1=\frac{1}{y-2} \\ \Rightarrow y-2=1 \\ \Rightarrow y=1+2=3 \end{gathered}$$

Therefore the values of $x=5andy=3$