Question

For what values of m does the system of equations $3x+my=m$ and $2x-5y=20$ has a solution satisfying the condition x > 0, y > 0.
The ans is $m\in (-\infty -\frac{15}{2})\cup (k,\infty )$
Find the value of k?

Answer 1

Given system of equations is
$3x+my=m$
$2x-5y=20$
Here $D=\left|\begin{array}{cc}3& m\\ 2& -5\end{array}\right|=-15-2m$
$D_1=\left|\begin{array}{cc}m& m\\ 20& -5\end{array}\right|=-25m$
and $D_2=\left|\begin{array}{cc}3& m\\ 2& 20\end{array}\right|=60-2m$
By Cramer's rule
$x=\frac{D_1}{D}=\frac{-25m}{-15-2m}=\frac{25m}{15+2m}$
and $y=\frac{D_2}{D}=\frac{60-2m}{-15-2m}=\frac{2m-60}{2m+15}$
But given x > 0
then $\frac{25m}{15+2m}>0$
From wavy curve method
$\therefore m\in (-\infty ,-\frac{15}{2})\cup (0,\infty )$
and y > 0
$\frac{2m-60}{2m+15}>0$
From wavy curve method
$m\in (-\infty ,-\frac{15}{2})\cup (30,\infty )$ (ii)
from (i) and (ii), we get
$m\in (-\infty -\frac{15}{2})\cup (30,\infty )$

$\therefore k=30$