Find the value of $k$ , for which the following system of linear equations has infinite number of solutions?
$2 x + 3 y = 7; (k + 1) x + (2 k - 1) y = 4 k + 1$

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Solution

Step 1: If the pair of equations has infinite number of solutions
then,
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Given,
$2 x + 3 y = 7 \Rightarrow 2 x + 3 y - 7 = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) (k + 1) x + (2 k - 1) y = (4 k + 1) \Rightarrow (k + 1) x + (2 k - 1) y - (4 k + 1) = 0 . . . . . . . . . . . . . . . . . . . (2)$
Equation (1) and (2) are in standard form
Step 2: Substituting the values in the condition $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$h e r e a_{1} = 2 b_{1} = 3 c_{1} = - 7 a_{2} = k + 1 b_{2} = 2 k - 1 c_{2} = - (4 k + 1) \Rightarrow \frac{2}{k + 1} = \frac{3}{2 k - 1} = \frac{- 7}{- (4 k + 1)}$
Step 3 :Solving the equation by cross multiplication method
$T a k i n g \frac{2}{k + 1} = \frac{3}{2 k - 1} \Rightarrow 2 (2 k - 1) = 3 (k + 1) \Rightarrow 4 k - 2 = 3 k + 3 \Rightarrow 4 k - 3 k = 3 + 2 \Rightarrow k = 5$
Therefore the value of $k = 5$

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Similar questions

2 x + 3 y = 7

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(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

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Find the value of $k$ for which the following system of linear equations has infinite solutions

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

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