Question

Greatest integral value of $m$ for which the system of equations $3x+my=m$ and $2x-5y=20$ has a solution satisfying $x>0,y<0$ is

• A. $28$
• B. $29$
• C. $30$
• D. does not exist

Answer 1

The correct option is B $29$

Given equations are $3x+my=m$ ..(1) and $2x-5y=20$ ..(2)

By cross multiplication method,

$\frac{x}{-20m-5m}=\frac{-y}{-60+2m}=\frac{1}{-15-2m}$

By solving $x$ and $y$ ,

$x=\frac{-\left(25m\right)}{-(2m+15)}=\frac{25m}{2m+15}$

similarly, $y=\frac{60-2m}{-(2m+15)}=\frac{2m-60}{2m+15}$

As per question, we need to have a solution of $x>0$ and $y<0$

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

for $y<0$ ,$\frac{2m-60}{2m+15}<0$

As $(2m+15)$ is common denominator; $(2m+15)>0$ and $25m>0$ and $(2m-60)(2m+15)<0$

$m>0$ and $(2m-60)<0$ as $2m+15>0$

$2m<60$

$m<30$

For $m>0$ and $m<30$; Maximum possible integral value of $m$ is $29$.