Question

Find the value of

k

, for the following system of equation has infinitely many solutions.

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

Accepted Answer

Given two linear equations are

$2 x - 3 y = 7$

$(k + 2) x - (2 k + 1) y = 3 (2 k - 1)$ $[\begin{array}{l} a_{1} x + b_{1} y + c_{1} = 0 \\ a_{2} x + b_{2} y + c_{2} = 0 \end{array}]$

Here $a_{1} = 2, b_{1} = - 3, a_{2} = (k + 2), b_{2} = - (2 k + 1), c_{1} = 7, c_{2} = 3 (2 k - 1)$

For infinite number of solution we consider $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{2}{k + 2} = \frac{- 3}{- (2 k + 1)} = \frac{7}{3 (2 k - 1)}$

$\frac{2}{k + 2} = \frac{- 3}{- (2 k + 1)}$ $\frac{- 3}{- (2 k + 1)} = \frac{7}{3 (2 k - 1)}$

$2 (2 k + 1) = 3 (k + 2)$ $9 (2 k - 1) = 7 (2 k + 1)$

$4 k + 2 = 3 k + 6$ $18 k - 9 = 14 k + 7$

$k = 4$ $4 k = 16$

$k = 4$

Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 56: Solved Example

Tamil Nadu Board Mathematics Solutions for Exercise - Tamil Nadu Board Solutions for Chapter: Algebra, Exercise 56: Solved Example

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