Question

7. For what value of k; x = 2, y = 3 is a solution of the equation (k+1)x - (2x + 3)y-1=0. Also write
the equation​

Answer 1

Answer:

ans : k=22

Step-by-step explanation:

x=2 y=3

(k+1) 2 - (2×2+3) 3 - 1 = 0

2k+2 - 21 -1 = 0

2k +2 - 21 =0+1

2k+2 = 1+21

2k =22 + 2

k =44/2

k = 22

Answer 2

Answer:     k = 10

Step-by-step explanation:

Given,

$(k+1)x-(2x+3)y-1=0$

Put x = 2 and y = 3,

$(k+1)(2)-[2(2)+3)](3)-1=0\\ (k+1)(2) - (7)(3)-1=0\\ (k+1)2-21-1=0\\ (k+1)2=22\\ (k+1)=22/2\\ (k+1)=11\\ k = 10$

The equation will be, 11x - 2xy - 3y = 1

$(10+1)x-(2x+3)y-1=0\\ 11x-(2x+3)y-1=0\\11x-2xy-3y-1=0\\ 11x-2xy-3y=1$