Question

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$$\begin{gathered} (a-1)x+3y=2, \\ 6x+(1-2b)y=6. \end{gathered}$$

Answer 1

The given system of equations:
(a − 1)x + 3y = 2
⇒ (a − 1)x + 3y − 2 = 0 ...(i)
and 6x + (1 − 2b)y = 6
⇒ 6x + (1 − 2b)y − 6= 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
where, a₁ = (a − 1), b₁= 3, c₁ = −2 and a₂ = 6, b₂ = (1 − 2b), c₂ = −6
For an infinite number of solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{\left(a-1\right)}{6}=\frac{3}{\left(1-2b\right)}=\frac{-2}{-6}$
$\Rightarrow \frac{a-1}{6}=\frac{3}{\left(1-2b\right)}=\frac{1}{3}$
$\Rightarrow \frac{a-1}{6}=\frac{1}{3}\mathrm{and}\frac{3}{\left(1-2\mathrm{b}\right)}=\frac{1}{3}$
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
⇒ a = 3 and b = −4
∴​ a = 3 and b = −4