For what value of k, the following system of linear equations has infinitely many solutions? 5x+2y=2k 2(k+1)x+ky=3k+4

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Solution

Given equations are:
5x+2y=2k
2(k+1)x+ky=3k+4

$\Rightarrow 5 x + 2 y - 2 k = 0 \Rightarrow 2 (k + 1) + k y - (3 k + 4) = 0 F o r t h e e q u a t i o n s t o h a v e i n f i n i t e l y m a n y s o l u t i o n s, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} H e r e, a_{1} = 5 a_{2} = 2 (k + 1) b_{1} = 2 b_{2} = k c_{1} = - 2 k c_{2} = - (3 k + 4) ∴ \frac{5}{2 (k + 1)} = \frac{2}{k} = \frac{- 2}{- (3 k + 4)} \Rightarrow \frac{5}{2 (k + 1)} = \frac{2}{k} \Rightarrow 5 k = 4 (k + 1) \Rightarrow 5 k = 4 k + 4 \Rightarrow k = 4$

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Similar questions

k

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}

3 x + 2 y = 1

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

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