For what value of 𝑝 will the following system of equations have no solution?

(2𝑝 − 1)𝑥 + (𝑝 − 1)𝑦 = 2𝑝 + 1; 𝑦 + 3𝑥 − 1 = 0

 

We need to find the value of 𝑝 for which
(2𝑝 − 1)𝑥 + (𝑝 − 1)𝑦 = 2𝑝 + 1; 𝑦 + 3𝑥 − 1 = 0 has no solution.
(2𝑝 − 1)𝑥 + (𝑝 − 1)𝑦 = 2𝑝 + 1 can also be written as

(2𝑝 − 1)𝑥 + (𝑝 − 1)𝑦 − (2𝑝 + 1) = 0. On comparing it with 𝑎₁ 𝑥 + 𝑏₁y + 𝑐₁ = 0,
we get
𝑎₁ = (2𝑝 − 1)
𝑏₁ = (𝑝 − 1)
𝑐₁ =− (2𝑝 + 1)
Also, for 𝑦 + 3𝑥 − 1 = 0, On comparing it with
𝑎₂ 𝑥 + 𝑏₂ y + c₂  = 0, we get
𝑎₂    = 3
𝑏₂   = 1
𝑐₂  =− 1
It is known that for a system of equations to have no solution, it must satisfy the following condition.

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

On substituting the values, we get

$$\begin{gathered} \frac{2p-1}{3}=\frac{p-1}{1}\ne \frac{-(2p+1)}{-1} \\ \frac{2p-1}{3}=\frac{p-1}{1} \end{gathered}$$

⇒ 2𝑝 − 1 = 3(𝑝 − 1)

⇒ 2𝑝 − 1 = 3𝑝 − 3

⇒ 𝑝 = 2

Now, on comparing last two terms, we get

$\frac{p-1}{1}\ne \frac{-(2p+1)}{-1}$

⇒ − (𝑝 − 1) ≠− (2𝑝 + 1)

⇒ 𝑝 − 1 ≠ 2𝑝 + 1

⇒ 𝑝 ≠− 2

Now, on comparing the first and last terms, we get

$\frac{2p-1}{3}\ne \frac{-(2p+1)}{-1}$

⇒ (2𝑝 − 1) ≠ 3(2𝑝 + 1)

⇒ 2𝑝 − 1 ≠ 6𝑝 + 3

⇒ 𝑝 ≠− 1

Hence, the system has no solution when 𝑝 = 2.