Question

Solve $23x-29y=98$And $29x-23y=110$

Answer 1

Consider

$$\begin{gathered} 23x–29y=98.....\left(1\right) \\ 29x–23y=110.....\left(2\right) \end{gathered}$$

Now by adding both equations $1$ and $2$, we get

$$\begin{gathered} 23x-29y+29x-23y=98+110 \\ 52x–52y=208 \\ 52(x–y)=208 \\ (x–y)=\frac{208}{52} \\ x–y=4.......\left(3\right) \end{gathered}$$

Now by subtracting $1$ from $2$ , we get

$$\begin{gathered} 29x-23y-\left(23x-29y\right)=110-98 \\ 29x-23y-23x+29y=12 \\ 6x+6y=12 \\ x+y=2......\left(4\right) \end{gathered}$$

Now by solving$\left(3\right)and\left(4\right),$ we get

$$\begin{gathered} x+y=2x-y=42x=6 \\ x=\frac{6}{2} \\ x=3 \end{gathered}$$

Now by substituting the value of $x$in $\left(4\right)$, we get

$$\begin{gathered} x+y=2 \\ 3+y=2 \\ y=2-3 \\ y=-1 \end{gathered}$$

Hence, $x=3andy=-1$