Question

The value(s) of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions

x > 0, y > 0

• A. MĪ (0, )
• B. M Ī$\left( –\infty,–\frac{15}{2} \right)$Č (30, )
• C. MĪ$\left( –\frac{15}{2},\infty \right)$
• D. None of these

Answer 1

Answer: (B)

M Ī$\left( –\infty,–\frac{15}{2} \right)$Č (30, )

Answer 2

By using Cramer’s rule, the solution of the system is

x = $\frac{\Delta_{x}}{\Delta}$, y = $\frac{\Delta_{y}}{\Delta}$, where D = $\left| \begin{matrix} 3 & m \\ 2 & - 5 \end{matrix} \right|$ = –(15 + 2m)

D_x = $\left| \begin{matrix} m & m \\ 20 & - 5 \end{matrix} \right|$ = –25m, D_y = $\left| \begin{matrix} 3 & m \\ 2 & 20 \end{matrix} \right|$ = 60 – 2m

Ž x = $\frac{- 25m}{–(15 + 2m)}$=$\frac{25m(15 + 2m)}{(15 + 2m)^{2}}$> 0,

for m > 0, or m < $–\frac{15}{2}$.

Also y = $\frac{60 - 2m}{- (15 + 2m)}$=$\frac{2(m - 30)(15 + 2m)}{(15 + 2m)^{2}}$>0

for m > 30, or m < $–\frac{15}{2}$.

Ž x > 0, y > 0 for m > 30 or m < $–\frac{15}{2}$

For m = $–\frac{15}{2}$, the system has no solution.

Hence (2) is correct answer.