Question

Find the values of a and b for which the following system of equations has infinitely many solutions.
$
  \left( {2a - 1} \right)x - 3y = 5 \\
  3x + \left( {b - 2} \right)y = 3 \\
 $

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 solution -
The given system of linear equations is
$
 \left( {2a - 1} \right)x - 3y = 5 \\
   \Rightarrow \left( {2a - 1} \right)x - 3y - 5 = 0{\text{ }} \to {\text{(1)}} \\ $
 $ and $
$ 3x + \left( {b - 2} \right)y = 3 \\
   \Rightarrow 3x + \left( {b - 2} \right)y - 3 = 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
${a_1} = 2a - 1,{b_1} = 3,{c_1} = - 5$
By comparing equations (2) and (4), we get
${a_2} = 3,{b_2} = b - 2,{c_2} = - 3$
For the given pair of linear equations to have infinitely many solutions, equation (5) must be satisfied
By equation (5), we can write
\[
  \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}} \\
   \Rightarrow \dfrac{{2a - 1}}{3} = \dfrac{3}{{b - 2}} = \dfrac{{ - 5}}{{ - 3}} \\
   \Rightarrow \dfrac{{2a - 1}}{3} = \dfrac{3}{{b - 2}} = \dfrac{5}{3}{\text{ }} \to {\text{(6)}} \\
 \]
By equation (6), we can write
\[
   \Rightarrow \dfrac{{2a - 1}}{3} = \dfrac{5}{3} \\
   \Rightarrow 2a - 1 = 5 \\
   \Rightarrow 2a = 5 + 1 = 6 \\
   \Rightarrow a = 3 \\
 \]
By equation (6), we can write
\[
   \Rightarrow \dfrac{3}{{b - 2}} = \dfrac{5}{3}{\text{ }} \\
   \Rightarrow 5\left( {b - 2} \right) = 3 \times 3 = 9 \\
   \Rightarrow 5b - 10 = 9 \\
   \Rightarrow 5b = 19 \\
   \Rightarrow b = \dfrac{{19}}{5} \\
 \]
Therefore, the required values of a and b for which the given system of linear equations has infinitely many solutions are 3 and \[\dfrac{{19}}{5}\] 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.