Question

For what values of m does the system of equations $3x+my=m$ and $2x-5y=20$ have a solution satisfying the conditions $x>0,y>0$.

Answer 1

The given equations are

$3x+my-m=0$

and

$2x-5y-20=0$

Solving these equations, we get

$x=\frac{25m}{2m+15},y=\frac{2m-60}{2m+15}$

For $x>0$

$\frac{25m}{2m+15}>0$

$\Rightarrow m<-\frac{15}{2}\text{or}m>0$ .......(1)

For $y>0$

$\frac{2(m-30)}{2m+15}>0$

$\Rightarrow m<-\frac{15}{2}$ or $m>30$ .......(2)

Combining (1) and (2), we get the common values of $m$

$m<-\frac{15}{2}\text{or}m>30$

$\therefore m\in (-\infty ,\frac{-15}{2})\cup (30,\infty )$