Find the value of $k$ for which the given system of equations has infinite number of solutions.
$5 x + 2 y = 2 k$ and $2 (k + 1) x + k y = (3 k + 4)$

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Solution

The correct option is A 4
Comparing $5 x + 2 y = 2 k$ with $a_{1} x + b_{1} y + c_{1} = 0$ , we get,
$a_{1} = 5, b_{1} = 2, c_{1} = - 2 k$

Comparing $2 (k + 1) x + k y = 3 k + 4$ with $a_{2} x + b_{2} y + c_{2} = 0$ , we get,
$a_{2} = 2 (k + 1), b_{2} = k, c_{2} = - (3 k + 4)$

Now,
Given: the equations has infinite no of solutions.
$∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{5}{2 (k + 1)} = \frac{2}{k} = \frac{- 2 k}{- (3 k + 4)}$

When,

$\frac{5}{2 (k + 1)} = \frac{2}{k}$

$⟹ 5 k = 4 (k + 1)$
$⟹ 5 k - 4 k = 4$
$⟹ k = 4$

and

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

$⟹ 2 (3 k + 4) = 2 k^{2}$
$⟹ 3 k + 4 = k^{2}$
$⟹ k^{2} - 3 k - 4 = 0$
$⟹ k^{2} - 4 k + k - 4 = 0$
$⟹ k (k - 4) + (k - 4) = 0$
$⟹ (k - 4) (k + 1) = 0$

Either, $k = 4 o r k = - 1$
$k = 4$ is the value that satisfies both the conditions.

So, $O p t i o n A i s c o r r e c t$

Suggest Corrections

Similar questions

5 x + 2 y = 2 k, 2 (k + 1) x + k y = (3 k + 4) .

5 x + 2 y = 2 k and 2 (k + 1) x + ky = (3 k + 4)

k = \frac{2}{3}

k = \frac{- 2}{3}

(2 k + 1) x + 5 k y = k + 2

k x + (k + 2) y = 2

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