The value(s) of $m$ does the system of equations $3 x + m y = m$ and $2 x - 5 y = 20$ has a solution satisfying the conditions $x > 0, y > 0$

m \in (0, \infty)

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

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m \in (\frac{15}{2}, \infty)

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None of these

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Solution

The correct option is B $m \in (\infty, - \frac{15}{2}) \cup (30, \infty)$
Given that

$\Rightarrow 3 x + m y = m$ ... $(1)$

$\Rightarrow 2 x - 5 y = 20$

$\Rightarrow x = \frac{20 + 5 y}{2}$ ... $(2)$

Substitute $(2)$ in $(1)$

$\Rightarrow \frac{60 + 15 y}{2} + m y = m$

On simplifying we get

$\Rightarrow y = \frac{2 m - 60}{15 + 2 m}$

Given that $x > 0; y > 0$

Hence, $\frac{2 m - 60}{15 + 2 m} > 0$

$\Rightarrow \frac{(2 m - 60) (15 + 2 m)}{(15 + 2 m)^{2}} > 0$

$\Rightarrow (2 m - 60) (15 + 2 m) > 0$

$\Rightarrow 4 (m - 30) (m + \frac{15}{2}) > 0$

Therefore, $m > 30; m < \frac{- 15}{2}$

Thus, $m$ lies in the interval, $(- \infty, - \frac{15}{2}) \cup (30, \infty)$

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