Question

For what value of k do the equations 3(k - 1)x + 4y = 24 and 15x + 20y = 8(k + 13) have infinite solutions?
1. 3
2. 4
3. 2
4. 1

Answer 1

Correct Answer - Option 3 : 2

Concept-

• In Infinite solution the lines are coincident.
• For a system of equations having infinite solution,

a1 x + b1 y = c1 and a2 x + b2 y = c2  

\(\text{we have}, \dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}= \dfrac{c_{1}}{c_{2}}\)

Calculation-

Given equation-

3(k - 1)x + 4y = 24

15x + 20y = 8(k + 13)

We can write 1st and 2nd equation as-

3(k - 1)x + 4y - 24 = 0      ---(i)

15x + 20y - 8(k + 13) = 0      ---(ii)

We know that in Infinite solution the lines are coincident.

∴ Its equation is -

a₁ x + b₁ y = c₁      ---(iii)

a₂ x + b₂ y = c₂      ---(iv)

\(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}= \dfrac{c_{1}}{c_{2}}\)     ---(v)

On comparing equation (i) with (iii) and (ii) with (iv), we get-

a₁ =  3(k - 1), a₂ = 15

b₁ = 4, b₂ = 20

c₁ = -24, c₂ = -8(k + 13) 

Substituting the value of a₁, a₂, b₁, b₂, c₁ and c₂ in equation (v), we get-

⇒   \(\dfrac{3(k-1)}{15}\) = \(\dfrac{4}{20}\) = \(\dfrac{-24}{-8(k+13)}\)

⇒  \(\dfrac{(k-1)}{5}\) = \(\dfrac{1}{5}\) 

⇒ k - 1 = 1

⇒ k = 2

Hence, At k = 2 the equations 3(k - 1)x + 4y = 24 and 15x + 20y = 8(k + 13) have infinite solutions.