value of p,x=2.y=3 is a solution of (p+1)x-(x2p+3)y-1=0
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hello users.......
we have to find the value of p=?
given that
x=2.y=3 is a solution of equation (p+1)x-(x²p+3)y-1=0
solution:-
putting the value
x=2.y=3 in equation, we get
=> (p+1)2 - (2²p +3)3 -1 = 0
= 2p +2 -(4p +3)3 -1 = 0
=. 2p +2 -12p -9 -1 = 0
=>. -10p -8 = 0
=>. p = -8/10 = -4/5 answer
❇❇ hope it helps ❇❇
we have to find the value of p=?
given that
x=2.y=3 is a solution of equation (p+1)x-(x²p+3)y-1=0
solution:-
putting the value
x=2.y=3 in equation, we get
=> (p+1)2 - (2²p +3)3 -1 = 0
= 2p +2 -(4p +3)3 -1 = 0
=. 2p +2 -12p -9 -1 = 0
=>. -10p -8 = 0
=>. p = -8/10 = -4/5 answer
❇❇ hope it helps ❇❇
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