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Maths-

General

Easy

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

m $\in$ (0, $\infty$ )
m $\in$ $(- \infty, - \frac{15}{2})$ $\cup$ (30, $\infty$ )
m $\in$ $(- \frac{15}{2}, \infty)$
None of these

The correct answer is: m $element of$ $open parentheses – infinity comma – fraction numerator 15 over denominator 2 end fraction close parentheses$ $union$ (30, $infinity$ )

By using Cramer’s rule, the solution of the system is
x = $fraction numerator capital delta subscript x end subscript over denominator capital delta end fraction$ , y = $fraction numerator capital delta subscript y end subscript over denominator capital delta end fraction$ , where $straight capital delta$ = $open vertical bar table row 3 m row 2 cell negative 5 end cell end table close vertical bar$
= –(15 + 2m)
$straight capital delta$ _x = $open vertical bar table row m m row 20 cell negative 5 end cell end table close vertical bar$ = –25m, $straight capital delta$ _y = $open vertical bar table row 3 m row 2 20 end table close vertical bar$ = 60 – 2m
$rightwards double arrow$ x = $fraction numerator negative 25 m over denominator – left parenthesis 15 plus 2 m right parenthesis end fraction$ = $fraction numerator 25 m left parenthesis 15 plus 2 m right parenthesis over denominator left parenthesis 15 plus 2 m right parenthesis to the power of 2 end exponent end fraction$ > 0,
for m > 0, or m < $– fraction numerator 15 over denominator 2 end fraction$ .
Also y = $fraction numerator 60 minus 2 m over denominator negative left parenthesis 15 plus 2 m right parenthesis end fraction$ = $fraction numerator 2 left parenthesis m minus 30 right parenthesis left parenthesis 15 plus 2 m right parenthesis over denominator left parenthesis 15 plus 2 m right parenthesis to the power of 2 end exponent end fraction$ >0
for m > 30, or m < $– fraction numerator 15 over denominator 2 end fraction$ .
$rightwards double arrow$ x > 0, y > 0 for m > 30 or m < $– fraction numerator 15 over denominator 2 end fraction$
For m = $– fraction numerator 15 over denominator 2 end fraction$ , the system has no solution.
Hence (B) is correct answer.

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