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) $m\in \left( 0,\infty \right)$
(b) $m\in \left( -\infty ,\dfrac{-15}{2} \right)\cup \left( 30,\infty \right)$
(c) $m\in \left( \dfrac{15}{2},\infty \right)$
(d) None of these

Answer 1

Hint: Solve the given linear equations using elimination method and then apply the condition that $x>0$ and $y>0$ on the calculated solutions to find the values of m that satisfy the given linear equations.

Complete step-by-step answer:
We have the system of linear equations $3x+my=m$ and $2x-5y=20$. We have to find the values of m which satisfy the system of equations for $x>0$ and $y>0$.
We will solve the linear equations by elimination method and then apply the conditions $x>0$ and $y>0$.
To solve the given equations, multiply the equation $3x+my=m$ by 5 and $2x-5y=20$ by m and add the two equations.
Thus, we have $\left( 15x+5my \right)+\left( 2mx-5my \right)=5m+20m$.
Simplifying the above equation, we have $x\left( 15+2m \right)=25m$.
Rearranging the terms, we have $x=\dfrac{25m}{15+2m}$.
Substituting the above value in equation $2x-5y=20$, we have $2\left( \dfrac{25m}{15+2m} \right)-5y=20$.
Simplifying the above equation, we have $\dfrac{50m}{15+2m}-20=5y$.
Rearranging the terms, we have $5y=\dfrac{50m-300-40m}{15+2m}$.
Thus, we have $5y=\dfrac{10m-300}{15+2m}\Rightarrow y=\dfrac{2m-60}{15+2m}$.
Now, we know that $x>0,y>0$.
Thus, we have $x=\dfrac{25m}{15+2m}>0,y=\dfrac{2m-60}{15+2m}>0$.
Firstly we observe that $15+2m\ne 0$. Thus, we have $m\ne \dfrac{-15}{2}$.
As $\dfrac{25m}{15+2m}>0$, we have $25m>0,15+2m>0$ or $25m<0,15+2m<0$.
Thus, we have $\left( m>0 \right)\cap \left( m>\dfrac{-15}{2} \right)$ or $\left( m<0 \right)\cap \left( m<\dfrac{-15}{2} \right)$.
So, we have $m>0$ or $m<\dfrac{-15}{2}$.
Similarly, as $\dfrac{2m-60}{15+2m}>0$, we have $2m-60>0,15+2m>0$ or $2m-60<0,15+2m<0$.
Thus, we have $\left( m>30 \right)\cap \left( m>\dfrac{-15}{2} \right)$ or $\left( m<30 \right)\cap \left( m<\dfrac{-15}{2} \right)$.
So, we have $m>30$ or $m<\dfrac{-15}{2}$.
So, the possible values of m are $\left( m>0 \right)\cap \left( m>30 \right)$ or $\left( m<\dfrac{-15}{2} \right)\cap \left( m<\dfrac{-15}{2} \right)$.
Thus, we have $m>30$ or $m<\dfrac{-15}{2}$.
Hence, the values of m which satisfy the given system of linear equations are $m\in \left( -\infty ,\dfrac{-15}{2} \right)\cup \left( 30,\infty \right)$, which is option (b).

Note: We can check if the calculated values of m satisfy the given system of linear equations or not by substituting the value of m in the system of equations. It’s necessary to use the fact that $x>0,y>0$; otherwise, we won’t be able to solve this question.