Question

determine the values of a and b so that the following system of linear equations have infinitely many solutions
(2a – 1)x + 3y – 5 = 0
3x + (b – 1)y – 2 = 0​

Answer 1

Answer:

${a_1}$ = 2a – 1

${a_2}$ = 3

${b_1}$ = 3

${b_2}$ = b – 1

${c_1}$ = – 5

${c_2}$ = – 2

${a_1}$/${a_2}$ = ${b_1}$/${b_2}$ = ${c_1}$/${c_2}$

${a_1}$/${a_2}$ = ${b_1}$/${b_2}$

$\Longrightarrow \frac{2a - 1}{3} = \frac{3}{b - 1}$

$(2a - 1)(b - 1) = 9 \\ \\ \Longrightarrow2ab - b - 2a +1 = 9 \\ \Longrightarrow \: b(2a - 1) = \frac{8 + 2a}{2a - 1} \: \: \: \: \: ...(1)$

$\Longrightarrow$ ${b_1}$/${b_2}$ = ${c_1}$/${c_2}$

Put, value of b in equation 1 from equation 2

$b = \frac{8 + 2a}{2a - 1} \\ \\ \Longrightarrow \: \frac{11}{5} = \frac{8 + 2a}{2a - 1} \\ \\ \Longrightarrow11(2a + 1) = 5(8 + 2a) \\ \\ \Longrightarrow22a - 11 = 40 + 10a \\ \\\Longrightarrow12a = 51 \\ \\ \Longrightarrow \: a = \frac{51}{12}$