The linear pairs of equations $(3 k + 1) x + 3 y - 5 = 0 a n d 2 x - 3 y + 5 = 0$ have infinite solutions. Find the value of $K .$

- 1

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

- 2

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

- 3

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

- 4

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

Open in App

Solution

The correct option is A $- 1$
The linear pairs of equations $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ have infinite solutions if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Consider the equations $(3 k + 1) x + 3 y - 5 = 0$ and $2 x - 3 y + 5 = 0$
$\Rightarrow \frac{3 k + 1}{2} = \frac{3}{- 3} = \frac{- 5}{5}$
$\Rightarrow \frac{3 k + 1}{2} = - 1$
$\Rightarrow 3 k + 1 = - 2$
$\Rightarrow 3 k = - 3$
$∴ k = \frac{- 3}{3} = - 1$
Hence the value of $k = - 1$

Suggest Corrections

Similar questions

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

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

(3 k + 1) x + 3 y - 2 = 0

(k^{2} + 1) x + (k - 2) y - 5 = 0

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

k

2 x + 3 y = 1

(3 k - 1) x + (1 - 2 k) y = 2 k + 3

Determine the value of k so that the following system of equation have no solution

(3k+1)x+3y-2=0

(K^2+1)x+(k-2)y-5=0

Join BYJU'S Learning Program