Question

Find the value of 'k' for which the following systems of linear equations have infinite number of solutions:
2x+ky=3
(k+5)x+25y=2k+5
​

Answer 1

$\large\sf {\orange {\underline{ \pink { \underline{\pink{Go \: through \: the \: above \: attachment }}}}}}$

Thank you!

@itzshivani

Answer 2

Given:

Pair of linear equation having infinite number of solution

2x+ky=3

(k+5)x+25y=2k+5

To find :

value of k

Solution:

$we \: \: know \: \: that \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ For \: \: finite \: \: solution \: , \: \: \: \: \: \: \: \: \\ \frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} \\ \\ \frac{2}{k + 5} = \frac{k}{25} = \frac{ - (3)}{ - (2k + 5)} \\ \\ \frac{2}{k + 5} = \frac{k}{25} \: \: \: or \: \: \: \frac{2}{k + 5} = \frac{3}{2k + 5} \\ \\ 50 = {k}^{2} + 5k \: \: \: or \: \: \: 4k + 10 = 3k + 15 \\ \\ {k}^{2} + 5k - 50 = 0 \: \: \: or \: \: \: 4k - 3k = 15 - 10 \\ \\ {k}^{2} + 10k - 5k - 50 = 0 \: \: \: or \: \: \: k = 5 \\ \\ k(k + 10) - 5(k + 10) = 0 \: \: or \: \: k = 5 \\ \\ (k + 10)(k - 5) = 0 \: \: or \: \: k = 5 \\ \\ k = 5, - 10 \: \: \: \: \: or \: \: \: \: \: k = 5 \\ \\ thus \: \: k = 5$