Question

for what value of k ',
$x = 2$
And
$y = 3$
is a solution of

$(k + 1)x - (2k + 3)y - 1 = 0$
also write the equation??? ¿? ¿? ¿? ¿? ¿? ​

Answer 1

Explanation:

$(k + 1)x - (2k + 3)y - 1 = 0 \\ x = 2 \\ y = 3 \\ putting \: values \: of \: x \: and \: y \\ (k + 1)2 - (2k + 3)3 - 1 = 0 \\ 2k + 2 - 6k - 9 - 1 = 0 \\ - 4k - 8= 0 \\ - 4k = 8 \\ k = - 2 \\ therefore \: required \: equation \: is \\ (- 2 + 1)x- (2 \times - 2 + 3)y - 1 = 0 \\ - x - y - 1 = 0 \\ - 1(x + y + 1) = 0 \\ x + y + 1 = 0$

Answer 2

Answer:

x+y+1=0

Explanation:

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