For what value of k, will the following pair of linear equations in two variable have infinitely many solutions?

$2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2$

k = 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

k = 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = 5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A k = 2
Given: $2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2$

Condition for infinitely many solutions is

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$a_{1} = 2, a_{2} = k + 2, b_{1} = 3, b_{2} = 6, c_{1} = - 4, c_{2} = - (3 k + 2)$

$\frac{2}{k + 2} = \frac{3}{6} = \frac{- 4}{- (3 k + 2)}$

$\frac{2}{k + 2} = \frac{3}{6}$

$\frac{2}{k + 2} = \frac{1}{2}$

$k + 2 = 4$

$k = 2$

Suggest Corrections

Similar questions

2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2

2 x + 3 y = 4

(k + 2) x + 6 y = 3 k + 2

2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2

For what value of k, will the following system of equations have infinitely many solutions?
$2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2$

Find the value of k for which the following pair of linear equations have infinitely many solutions:

$2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k .$

Join BYJU'S Learning Program