Question

for what value of p x=2,y=3 is a solution of (p+1)x-(2p+3)y-1=0 and write the equation .

Answer 1

(p+1)x-(2p+3)y-1=0
x = 2
y = 3


(p+1)x-(2p+3)y-1=0
px + x -( 2py +3y)  -1 =0
px + x -2py - 3y -1= 0
putting the value of x and y 
p(2) + 2 - 2p(3) - 3(3) -1 = 0
2p  +2 -6p  -9  -1 = 0
-4p  - 8 = 0
-4p = 8
p = -2

Answer 2

Answer:

Required value of p is $\frac{5}{2}$

Step-by-step explanation:

Given equation is $(p+1)x-(2p+3)y-1=0.......(1)$

and it is given that value of x is 2 and value of y is 3.

Here we want to find value of p.

We are putting x=2 and y=3 in equation number (1),

$(p+1) \times 2-(2p+3) \times 3-1=0$

First we are removing first brackets,

$2p+2-6p+9-1=0$

Now we are separating variable and constant part,

$2p - 6p = - 2 - 9 + 1$

$- 4p = - 10$

$p = \frac{10}{4} = \frac{5}{2}$

Required value of p is $\frac{5}{2}$