Question

Solve in positive integers:
$20x-21y=38$
$3y+4z=34$.

Answer 1

$20x-21y=38⟶1$

$3y+4z=34⟶2$

adding equation 1 and equation 2

$20x-18y+4z=72⟶3$

From Equation 1

$21y=20x-38$

$\therefore y=\frac{20x-38}{21}$

Now, putting this value in equation 2

$3\left(\frac{73-7z}{4}\right)+4z=34$

$⟹60x-114+84z=714$

$84z=828-60x$

$⟹z=\frac{828-60x}{84}$

$z=\frac{4(207-15x)}{84}$

$z=\frac{207-15x}{21}$

Now, putting $z=7,$$x=4$$y=2,x=4,z=7$