For k=, the following system of equations has infinite solutions: 4x + 2y - (k + 2) = 0 and 2x + y - 3 = 0

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

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

Open in App

Solution

The correct option is C 4
Given two equations $a_{1} x + b_{1} y = c_{1} a n d a_{2} x + b_{2} y = c_{2},$ the pair of equations will have infinite solutions if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}},$

Thus, for the given pair of equations to have infinte solutions,
$\frac{4}{2} = \frac{2}{1} = \frac{- (k + 2)}{- 3}$
$\Rightarrow \frac{(k + 2)}{3} = 2$
$\Rightarrow k = 4$

Suggest Corrections

Similar questions

k

2 x - k y + 3 = 0

3 x + 2 y - 1 = 0

Write the value of $k$ for which the system of equations $3 x - 2 y = 0$ and $k x + 5 y = 0$ has infinitely many solutions.

2 x - y + z = 0

x - 2 y + z = 0

λ x - y + 2 z = 0

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