Question

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

Answer 1

Your question must be this:

Find k for which the system of equation has infinitely many solutions 2x+(k-2)y=k and 6x+(2k-1)y=2k+5

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Answer 2

Answer:

k= 5

Step-by-step explanation:

Given : 2x + ( k - 2 )y =k ; 6x + ( 2k - 1 )y =( 2k +5)

from many solution

$\frac{a1}{a2} = \frac{b1}{b2} =\frac{c1}{c2}\\ = \frac{2}{6}=\frac{k - 2}{2k - 1} = \frac{k}{2k + 5} \\\frac{2}{6} =\frac{k - 2}{2k - 1} \\ 4k -2 = 6k - 12 \\ 4k - 6k = -12 +2 \\ -2k = -10 \\ k =\frac{-10}{-2}\\ k = 5\\\\now, \frac{2}{6}=\frac{k}{2k + 1} \\ 4k +10 = 6k \\ 2k = 10 \\k= \frac{10}{2}\\k = 5\\ \\ hence k = 5$