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.