Question

Determine the values of a and b, so that the following system of linear equations have infinitely many solutions.
 $
  \left( {2a - 1} \right)x + 3y - 5 = 0 \\
  3x + \left( {b - 1} \right)y - 2 = 0 \\
 $

Answer 1

Hint- Here, we will proceed by comparing the given pair of linear equations with any general pair of linear equations i.e., ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$. Then using the condition for having infinitely many solutions i.e., $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}$.
Complete step by step answer:
The given system of linear equations are $\left( {2a - 1} \right)x + 3y - 5 = 0{\text{ }} \to {\text{(1)}}$ and $3x + \left( {b - 1} \right)y - 2 = 0{\text{ }} \to {\text{(2)}}$
As we know that for any pair of linear equations ${a_1}x + {b_1}y + {c_1} = 0{\text{ }} \to {\text{(3)}}$ and ${a_2}x + {b_2}y + {c_2} = 0{\text{ }} \to {\text{(4)}}$ to have infinitely many solutions, the condition which must be satisfied is that the ratio of the coefficients of x should be equal to the ratio of the coefficients of y which further should be equal to the ratio of the constant terms in the pair of linear equations.
The condition is $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}{\text{ }} \to (5{\text{)}}$
By comparing equations (1) and (3), we get
${a_1} = \left( {2a - 1} \right),{b_1} = 3,{c_1} = - 5$
By comparing equations (2) and (4), we get
${a_2} = 3,{b_2} = \left( {b - 1} \right),{c_2} = - 2$
For the given pair of linear equations to have inconsistent solution, equation (5) must be satisfied
By equation (5), we can write
\[
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{{\left( {2a - 1} \right)}}{3} = \dfrac{3}{{\left( {b - 1} \right)}} = \dfrac{{ - 5}}{{ - 2}} \\
   \Rightarrow \dfrac{{2a - 1}}{3} = \dfrac{3}{{b - 1}} = \dfrac{5}{2}{\text{ }} \to {\text{(6)}} \\
 \]
By equation (6), we can write
\[
   \Rightarrow \dfrac{{2a - 1}}{3} = \dfrac{5}{2} \\
   \Rightarrow 2a - 1 = 3 \times \left( {\dfrac{5}{2}} \right) = \dfrac{{15}}{2} \\
   \Rightarrow 2a = \dfrac{{15}}{2} + 1 = \dfrac{{15 + 2}}{2} = \dfrac{{17}}{2} \\
   \Rightarrow a = \dfrac{{17}}{4} \\
 \]
By equation (6), we can write
\[
   \Rightarrow \dfrac{3}{{b - 1}} = \dfrac{5}{2}{\text{ }} \\
   \Rightarrow b - 1 = 3 \times \left( {\dfrac{2}{5}} \right) = \dfrac{6}{5} \\
   \Rightarrow b = \dfrac{6}{5} + 1 = \dfrac{{6 + 5}}{5} = \dfrac{{11}}{5} \\
 \]
Therefore, the required values of a and b for which the given system of linear equations has infinitely many solutions are \[\dfrac{{17}}{4}\] and \[\dfrac{{11}}{5}\] respectively.
Note- Any general pair of linear equations which are given by ${a_1}x + {b_1}y + {c_1} = 0$ and ${a_2}x + {b_2}y + {c_2} = 0$ can also have unique solution (consistent solution) if the condition $\dfrac{{{a_1}}}{{{a_2}}} \ne \dfrac{{{b_1}}}{{{b_2}}}$ is satisfied. Also, for these pair of linear equations to have no solution, the condition $\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} \ne \dfrac{{{c_1}}}{{{c_2}}}$ should always be satisfied.