Question

Solve in positive integers:
$13x+11z=103$
$7z-5y=4$.

Answer 1

$13x+11z=103⟶1$

$7z-5y=4⟶2$

From Equation 1

$11z=103-13x$

$\therefore z=\frac{103-13x}{11}$

Now, putting this value in equation 1

$7\left(\frac{103-13x}{11}\right)-5y=4$

$⟹721-91x-55y=44$

$55y=-91x+677$

$⟹y=\frac{677-91x}{55}$

Now, putting $x=2,$

$y=\frac{677-182}{55}=\frac{495}{55}=9$

$\therefore y=9$

and $z=\frac{103-26}{11}=\frac{77}{11}$

$z=7$

$\therefore x=2,y=9,z=7$