Question

Solve for $xandy$

$$\begin{gathered} \frac{24}{2x+y}-\frac{13}{3x+2y}=2 \\ \frac{26}{3x+2y}+\frac{8}{2x+y}=3 \end{gathered}$$

Answer 1

Step 1:Giving substitution

Given ,

$$\begin{gathered} \frac{24}{2x+y}-\frac{13}{3x+2y}=2 \\ \frac{26}{3x+2y}+\frac{8}{2x+y}=3 \end{gathered}$$

$$\begin{gathered} A=\frac{1}{2x+y}\&B=\frac{1}{3x+2y} \\ Therefore,24A-13B=2.......................................\left(1\right) \\ 8A+26B=3.........................................\left(2\right) \end{gathered}$$

Step 2 : Finding A and B using elimination method

$$\begin{gathered} equation\left(1\right)\times 2 \\ \Rightarrow 48A-26B=4.....................................\left(3\right) \\ 8A+26B=3.............................................\left(2\right) \\ \left(3\right)+\left(2\right)\Rightarrow 56A=7 \\ \Rightarrow A=\frac{7}{56}=\frac{1}{8} \\ substituteA=\frac{1}{8}inequation\left(2\right) \\ equation\left(2\right)\Rightarrow 8x\frac{1}{8}+26B=3 \\ \Rightarrow 1+26B=3 \\ \Rightarrow 26B=2 \\ B=\frac{2}{26}=\frac{1}{13} \end{gathered}$$

Step 3 : Finding $xandy$

We have ,

$$\begin{gathered} A=\frac{1}{2x+y}\Rightarrow \frac{1}{8}=\frac{1}{2x+y} \\ \Rightarrow 2x+y=8..................................\left(4\right) \\ B=\frac{1}{3x+2y}\Rightarrow \frac{1}{13}=\frac{1}{3x+2y} \\ \Rightarrow 3x+2y=13..........................\left(5\right) \\ Solving\left(4\right)\&\left(5\right) \\ Equation\left(4\right)\times 2\Rightarrow 4x+2y=16............................\left(6\right) \\ 5\Rightarrow 3x+2y=13..............................\left(5\right) \\ 6-5\Rightarrow x=3 \\ Substitutex=3inequation\left(4\right) \\ 4\Rightarrow 6+y=8 \\ \Rightarrow y=2 \end{gathered}$$

Therefore the values of $x=3andy=2$