For what value of $k$ , the following system of linear equations has infinite number of solutions
$2 x + 3 y = 2, (k + 2) x + (2 k + 1) y = 2 (k - 1)$

Open in App

Solution

Step 1: If the pair of linear equations has infinite number of solutions
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Given,
$2 x + 3 y - 2 = 0 . . . . . . . . . . (1) (k + 2) x + (2 k + 1) y - 2 (k - 1) = 0 . . . . . . . . . (2) here a_{1} = 2, b_{1} = 3, c_{1} = - 2 a_{2} = k + 2, b_{2} = 2 k + 1, c_{2} = - 2 (k - 1)$
$\Rightarrow \frac{2}{k + 2} = \frac{3}{2 k + 1} = \frac{- 2}{- 2 (k - 1)}$
Step 2: Cross Multiplying the equation
$Taking \frac{2}{k + 2} = \frac{3}{2 k + 1} \Rightarrow 2 (2 k + 1) = 3 (k + 2) \Rightarrow 4 k + 2 = 3 k + 6 \Rightarrow 4 k - 3 k = 6 - 2 \Rightarrow k = 4$
Step 4:Substituting the value of $k = 4$ and check ratio
We get
$\frac{2}{k + 2} = \frac{2}{6} = \frac{1}{3} \frac{3}{2 k + 1} = \frac{3}{9} = \frac{1}{3} \frac{- 2}{- 2 (k - 1)} = \frac{1}{3}$
Therefore the value of $k = \frac{1}{3}$

Suggest Corrections

Similar questions

2 x + (k - 2) y = k, 6 + (2 k - 1) y = (2 k + 5) .

2 x + 3 y = 7

(a - b) x + (a + b) y = 3 a + b - 2

3 x + y = 1

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

For what value of $k$ will the following system of linear equations has infinite number of solutions

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

3 x + y = 1

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

Join BYJU'S Learning Program