Question

Solve:

$\begin{array}{c}23x-29y=98\\ 29x-22y=110\end{array}$

Answer 1

$23x-29y=98$ ........$\left(1\right)$

$29x-22y=110$ ........$\left(2\right)$

From $\left(1\right)\Rightarrow x-\frac{29}{23}y=\frac{98}{23}$ ........$\left(3\right)$

From $\left(2\right)\Rightarrow x-\frac{22}{29}y=\frac{110}{29}$ ........$\left(4\right)$

Equations $\left(3\right)-\left(4\right)$

$\Rightarrow x-\frac{29}{23}y-x+\frac{22}{29}y=\frac{98}{23}-\frac{110}{29}$

$\Rightarrow (-\frac{29}{23}+\frac{22}{29})y=\frac{98}{23}-\frac{110}{29}$

$\Rightarrow y=\frac{\frac{98}{23}-\frac{110}{29}}{-\frac{29}{23}+\frac{22}{29}}=\frac{2842-2530}{-841+506}=-\frac{312}{335}$

Put $y=-\frac{312}{335}$ in $\left(1\right)$ we get

$23x-29\times -\frac{312}{335}=98$

$\Rightarrow 23x=98-\frac{9048}{335}=\frac{32830-9048}{335}=\frac{23782}{335}$

$\therefore x=\frac{23782}{335\times 23}=\frac{1034}{335}$

$\therefore x=\frac{1034}{335},y=-\frac{312}{335}$