Question

If $4x+3y=120$, find how many positive integer solutions are possible?

Answer 1

$3y+4x=120$

$y=\frac{120-4x}{3}$

$y=4\left(\frac{30-x}{3}\right)$

$=40-\frac{4x}{3}$

For $y$ to be positive integer, $(40-\frac{4x}{3})>0$

And $x$ should be multiple of $3$

$\frac{4x}{3}<40$ $\Rightarrow x<30$

$x$ is also positive integer

$\therefore x>0$

For $y$ to be positive integer, $x=3k$ and $0<x<3k$

$\therefore$ Solution set $=(3k,40,4k),k=1,\mathrm{2....9}$

$9$ solution are possible