Question

Find the values of \[m\] and \[n\] for which the following system of linear equations has infinitely many solutions.
\[3x + 4y = 12\] and \[\left( {m + n} \right)x + 2\left( {m - n} \right)y = 5m - 1\].

Answer 1

Hint:
Here we have to use the concept of the system of linear equations to get the values of\[m\] and \[n\]. We will use the condition for the infinitely many solutions to find the two equations. Then we will solve the equations to get the value of \[m\] and \[n\]. Linear equation is a type of equation which has a highest degree of 2.

Complete Step by Step Solution:
Given equations are \[3x + 4y = 12\] and \[\left( {m + n} \right)x + 2\left( {m - n} \right)y = 5m - 1\].
Now we will use the basic condition of the system of linear equations which has infinitely many solutions.
So for the equations \[{a_1}x + {b_1}y = {c_1}\] and \[{a_2}x + {b_2}y = {c_2}\], condition for the solution to be infinite many solutions, we get \[\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\].
Now by applying this condition to the given equation, we get
\[\dfrac{3}{{m + n}} = \dfrac{4}{{2\left( {m - n} \right)}} = \dfrac{{12}}{{5m - 1}}\]
Now by using this we will for the two equation and then we will solve it to get the value of \[m\] and \[n\]. Therefore, we get
\[\dfrac{3}{{m + n}} = \dfrac{4}{{2\left( {m - n} \right)}}\]
On cross multiplying the terms, we get
\[ \Rightarrow 3 \times 2\left( {m - n} \right) = 4\left( {m + n} \right)\]
Multiplying the terms, we get
\[ \Rightarrow 6m - 6n = 4m + 4n\]
Adding and subtracting the terms, we get
\[ \Rightarrow 2m - 10n = 0\]………………………… \[\left( 1 \right)\]
Now we will take the condition of
\[\dfrac{3}{{m + n}} = \dfrac{{12}}{{5m - 1}}\]
On cross multiplying the terms, we get
\[ \Rightarrow 3\left( {5m - 1} \right) = 12\left( {m + n} \right)\]
Multiplying the terms, we get
\[ \Rightarrow 15m - 3 = 12m + 12n\]
Adding and subtracting the terms, we get
\[ \Rightarrow 3m - 12n = 3\]…………………………\[\left( 2 \right)\]
Multiplying equation \[\left( 1 \right)\] by 3, we get
\[\left( {2m - 10n} \right) \times 3 = 0 \times 3\]
\[ \Rightarrow 6m - 30n = 0\]………………….\[\left( 3 \right)\]
Multiplying equation \[\left( 2 \right)\] by 2, we get
\[\left( {3m - 12n} \right) \times 2 = 3 \times 2\]
\[ \Rightarrow 6m - 24n = 6\]………………...\[\left( 4 \right)\]
Now by subtracting the equation \[\left( 4 \right)\] from the equation \[\left( 3 \right)\] to get the value of the \[m\] and \[n\]. Therefore, we get
\[\begin{array}{l}6m - 30n - \left( {6m - 24n} \right) = 0 - 6\\ \Rightarrow 6m - 30n - 6m + 24n = - 6\end{array}\]
Subtracting the like terms, we get
\[ \Rightarrow 0 - 6n = - 6\]
\[ \Rightarrow - 6n = - 6\]
Dividing both side by \[ - 6\], we get
\[ \Rightarrow n = \dfrac{{ - 6}}{{ - 6}} = 1\]
Now putting the value of \[n\] in equation \[\left( 1 \right)\] to get the value of \[m\]. Therefore, we get
\[ \Rightarrow 2m - 10\left( 1 \right) = 0\]
\[ \Rightarrow 2m = 10\]
\[ \Rightarrow m = \dfrac{{10}}{2} = 5\]

Hence the value of \[m\] and \[n\] is \[m = 5\] and \[n = 1\].

Note:
We should know the condition of the infinitely many solutions of the non-homogeneous system of equations. Condition for the infinitely many solutions of the homogeneous system of equations is that the determinant of the coefficient matrix of the equation is equal to zero. But if the determinant of the coefficient matrix is not equal to 0 then the system of equations will have a unique solution.
A variable of an equation can have both the values i.e. real value and the imaginary values like in our example we get one real value.