Question

For which value of p, following pair of linear equations: px+y=p^2 and x+py=1, have no solution.

Answer 1

Answer:

p=1

Step-by-step explanation:

hope you understand it easily.....

Answer 2

Given : x+py=1 , px+y= p² has no solutions.​

To Find : Value of p

Solution:

Pair of linear equations

a₁x  +  b₁y + c₁  =  0

a₂x  +  b₂y + c₂  =  0

Consistent

if a₁/a₂ ≠ b₁/b₂   (unique solution  and lines intersects each others)

  a₁/a₂ = b₁/b₂ = c₁/c₂   (infinite solutions and line coincide each other )

Inconsistent

if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂  ( No solution , lines are parallel to each other)

x+py=1

px+y= p²

no solutions.​

=> 1/p  = p/1  ≠ 1/p²

1/p = p/1  =>  p² = 1  => p  = ±1

 p = 1

=> 1/1 = 1/1  =1/1   infinite solution

1/-1  = -1/1  ≠  1/(-1)²  no solution

Hence value of p  is  -1   for no solution

x  - y = 1

-x + y = 1

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