Matrices and Determinants 4 Question 30
31. For what values of $m$, does the system of equations $3 x+m y=m$ and $2 x-5 y=20$ has a solution satisfying the conditions $x>0, y>0$ ?
(1979, 3M )
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Answer:
Correct Answer: 31. (1)
Solution:
- Given system of equations are
$ 3 x+m y=m \quad \text { and } \quad 2 x-5 y=20 $
Here,
$ \begin{aligned} & \Delta=\left|\begin{array}{cc} 3 & m \\ 2 & -5 \end{array}\right|=-15-2 m \\ & \Delta _x=\left|\begin{array}{cc} m & m \\ 20 & -5 \end{array}\right|=-25 m \\ & \Delta _y=\left|\begin{array}{cc} 3 & m \\ 2 & 20 \end{array}\right|=60-2 m \end{aligned} $
If $\Delta=0$, then system is inconsistent, i.e. it has no solution.
If $\Delta \neq 0$, i.e. $m \neq \frac{15}{2}$, the system has a unique solution for any fixed value of $m$.
We have, $\quad x=\frac{\Delta _x}{\Delta}=\frac{-25 m}{-15-2 m}=\frac{25 m}{15+2 m}$
and
$ y=\frac{\Delta _y}{\Delta}=\frac{60-2 m}{-15-2 m}=\frac{2 m-60}{15+2 m} $
For $x>0, \frac{25 m}{15+2 m}>0$
$\Rightarrow m>0$
or $m<-\frac{15}{2}$
and $y>0, \frac{2 m-60}{2 m+15}>0 \Rightarrow m>30$ or $m<-\frac{15}{2}$
From Eqs. (i) and (ii), we get $m<-\frac{15}{2}$ or $m>30$
