Question

For which real values of \[m\] are the simultaneous equations \[y = mx + 3\] , \[y = \left( {2m - 1} \right)x + 4\] satisfied by at least one pair of real numbers \[\left( {x,y} \right)\] \[?\]

Answer 1

Hint: We solve the given system of the equations in two ways. The first method is solving the given system of the equations by substitution method. Which is the best method to solve the system of two linear equations with two unknowns.

Complete step by step solution:
Substitution method: Suppose \[ax + by = u\] , \[cx + dy = v\] be a system of two linear equations.
We can solve them step by step by substitution method. By substitution, replace one variable say \[y\] from one equation to another equation and find the value of another variable i.e., \[x\] . Substitute the variable value \[x\] in the first equation to find the value of \[y\] .
Given \[y = mx + 3\] --(1) and \[y = \left( {2m - 1} \right)x + 4\] --(2).
From the equation (1) put \[y = mx + 3\] in the equation (2), we get
 \[mx + 3 = \left( {2m - 1} \right)x + 4\]
 \[ \Rightarrow \] \[\left( {1 - m} \right)x = 1\]
 \[ \Rightarrow \] \[x = \dfrac{1}{{\left( {1 - m} \right)}}\] ---(3)
Substitute the value of \[x = \dfrac{1}{{\left( {1 - m} \right)}}\] in the equation (1), we get
 \[y = \dfrac{m}{{\left( {1 - m} \right)}} + 3\] \[ \Rightarrow y = \dfrac{{3 - 2m}}{{\left( {1 - m} \right)}}\] --(4)
From the equations (3) and (4) it is clear that the values of \[x\] and \[y\] exist if \[m \ne 1\] .
Hence, for all real values \[m\] such that \[m \ne 1\] are the simultaneous equations \[y = mx + 3\] , \[y = \left( {2m - 1} \right)x + 4\] satisfied by at least one pair of real numbers \[\left( {x,y} \right)\] .

Note: Note that another method is a matrix method by considering the corresponding augmented matrix and find the rank by row reduced echelon form. The system is consistent if the rank of the coefficient matrix is equal to the rank of the augmented matrix.