Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$k x + 3 y = (2 x + 1), 2 (k + 1) x + 9 y = (7 k + 1) .$

Open in App

Solution

The given system of equations:
kx + 3y = (2k + 1)
⇒ kx + 3y − (2k + 1) = 0 ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y − (7k + 1) = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁= 3, c₁ = −(2k + 1) and a₂ = 2(k + 1), b₂= 9, c₂ = −(7k + 1)
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$i . e . \frac{k}{2 (k + 1)} = \frac{3}{9} = \frac{- (2 k + 1)}{- (7 k + 1)}$
$\Rightarrow \frac{k}{2 (k + 1)} = \frac{1}{3} = \frac{(2 k + 1)}{(7 k + 1)}$

Now, we have the following three cases:
Case I:
$\frac{k}{2 (k + 1)} = \frac{1}{3}$
$\Rightarrow 2 (k + 1) = 3 k$
$\Rightarrow 2 k + 2 = 3 k$
$\Rightarrow k = 2$

Case II:
$\frac{1}{3} = \frac{2 k + 1}{7 k + 1}$
$\Rightarrow (7 k + 1) = 6 k + 3$
$\Rightarrow k = 2$

Case III:
$\frac{k}{2 (k + 1)} = \frac{2 k + 1}{7 k + 1}$
$\Rightarrow k (7 k + 1) = (2 k + 1) \times 2 (k + 1)$
$\Rightarrow 7 k^{2} + k = (2 k + 1) (2 k + 2)$
$\Rightarrow 7 k^{2} + k = 4 k^{2} + 4 k + 2 k + 2$
$\Rightarrow 3 k^{2} - 5 k - 2 = 0$
$\Rightarrow 3 k^{2} - 6 k + k - 2 = 0$
$\Rightarrow 3 k (k - 2) + 1 (k - 2) = 0$
$\Rightarrow (3 k + 1) (k - 2) = 0$
$\Rightarrow k = 2 or k = \frac{- 1}{3}$
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.

Suggest Corrections

Similar questions

k x + 3 y = 2 k + 1 2 (k + 1) x + 9 y = 7 k + 1

x + (k + 1) y = 5

(k + 1) x + 9 y = 8 k - 1

Find the values of a and b for which each of the following systems of linear equations has an infinite number of equations:

$2 x + 3 y = 7, (a + b + 1) x + (a + 2 b + 2) y = 4 (a + b) + 1.$

(a - 1) x + 3 y = 2, 6 x + (1 - 2 b) y = 6 .

2 x - 3 y = 7 (k + 1) x + (1 - 2 k) y = (5 x - 4)

Join BYJU'S Learning Program