Question

Determine the values of m and n so that the following system of linear equations have infinite number of solutions:
\[\begin{array}{*{35}{l}}
   \left( 2m\text{ }-1 \right)x\text{ + }3y\text{ }-\text{ }5\text{ }=\text{ }0 \\
   3x\text{ + }\left( n\text{ }-\text{ }1 \right)y\text{ }-\text{ }2\text{ }=\text{ }0 \\
\end{array}\]

Answer 1

Hint: Apply the condition applicable for infinite solutions in the system of linear equations.
So just need to solve the equations\[\dfrac{\left( 2m\text{ }-1 \right)}{3}=\dfrac{3}{\left( n\text{ }-\text{ }1 \right)}=\dfrac{5}{2}\], to get m and n.
Complete step by step solution:
In the question, we have to find the values of m and n so that the following system of linear equations has an infinite number of solutions. Now, we know that when the two equations are of the form: \[{{a}_{1}}x\text{ + }{{\text{b}}_{1}}y-{{c}_{1}}\text{ }=\text{ }0\,\,\,\,\And \,\,\,\,{{a}_{2}}x\text{ + }{{\text{b}}_{2}}y-{{c}_{2}}\text{ }=\text{ }0\,\,\], then the condition for x and y to have infinite solutions will be when we have:
\[\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{\text{b}}_{1}}}{{{\text{b}}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}\].
So here we will compare the equation \[\left( 2m\text{ }-1 \right)x\text{ + }3y\text{ }-\text{ }5\text{ }=\text{ }0\] with \[{{a}_{1}}x\text{ + }{{\text{b}}_{1}}y-{{c}_{1}}\text{ }=\text{ }0\,\] and we get: \[{{a}_{1}}=\left( 2m\text{ }-1 \right)\], \[{{b}_{1}}=3\] and \[{{c}_{1}}=5\]. Similarly, we will compare equation \[3x\text{ + }\left( n\text{ }-\text{ }1 \right)y\text{ }-\text{ }2\text{ }=\text{ }0\] with \[\,{{a}_{2}}x\text{ + }{{\text{b}}_{2}}y-{{c}_{2}}\text{ }=\text{ }0\,\,\]and we get: \[{{a}_{2}}=3\], \[{{b}_{2}}=(n-1)\] and \[{{c}_{2}}=2\]. Now we have to just apply the condition of the infinite solution and that is given as follows:
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{\text{b}}_{1}}}{{{\text{b}}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}} \\
 & \Rightarrow \dfrac{\left( 2m\text{ }-1 \right)}{3}=\dfrac{3}{\left( n\text{ }-\text{ }1 \right)}=\dfrac{5}{2} \\
\end{align}\]
So here we will first solve the equation \[\dfrac{\left( 2m\text{ }-1 \right)}{3}=\dfrac{5}{2}\] to get the value of m. So we have:
\[\begin{align}
  & \Rightarrow \dfrac{\left( 2m\text{ }-1 \right)}{3}=\dfrac{5}{2} \\
 & \Rightarrow \left( 2m\text{ }-1 \right)=\dfrac{15}{2} \\
 & \Rightarrow m=\dfrac{17}{4} \\
\end{align}\]
So this gives the value of \[m=\dfrac{17}{4}\]. Next, we will solve the equation \[\dfrac{3}{\left( n\text{ }-\text{ }1 \right)}=\dfrac{5}{2}\], to get the value of n. So we will have:
\[\begin{align}
  & \Rightarrow \dfrac{3}{\left( n\text{ }-\text{ }1 \right)}=\dfrac{5}{2} \\
 & \Rightarrow 6=5\left( n\text{ }-\text{ }1 \right) \\
 & \Rightarrow 11=5\left( n \right) \\
 & \Rightarrow n=\dfrac{11}{5} \\
\end{align}\]
So from this equation we get the value \[n=\dfrac{11}{5}\]. Finally, the value of \[m=\dfrac{17}{4}\] and \[n=\dfrac{11}{5}\].

Note: The system of linear equations can be plotted using the table of values in the Cartesian plane. The overlapping lines will have infinite solutions. Whereas the set of linear equations that has the intersecting lines will have just one unique solution of x and y.