Question

Solve the following systems of equations:
$23x-29y=98$
$29x-23y=110$

Answer 1

$23x-29y=98⟶\left(i\right)29x-23y=110⟶\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$ we get $,$

$52x-52y=208\Rightarrow 52(x-y)=208\Rightarrow x-y=4⟶\left(iii\right)$

Substracting $\left(ii\right)$ from $\left(i\right),$

$-6x-6y=-12\Rightarrow x+y=2⟶\left(iv\right)$

Now by solving $\left(iii\right)$ and $\left(iv\right),$

$2x=6$

$\Rightarrow x=3$

Substituting value of $x$ in equation $\left(iv\right),$

$3+y=2$

$y=-1$

$\therefore x=3$ and $y=-1$