Question

For what positive values of p the linear equations px + 3y -(p-3)=0 and 12x + py - p=0 have infinity many solutions

Answer 1

HI !

 px + 3y - ( p-3)=0

a₁ = p , b₁ = 3 , c₁ = - (p-3)

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12x + py - p=0

a₂ = 12  , b₂ = p , c₂ = -p

since , the equations have infinite solutions , 

a₁/a₂ = b₁/b₂ = c₁/c₂

p/12 = 3/p =  p-3/p


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p/12 = 3/p

cross multiply ,

p² = 36

p = √36
 p = 6

for the value of p = 6 , the equations will have infinitely many solutions

Answer 2

Hi Friend,

Here is your answer,

px + 3y - ( p-3)=0

a₁ = p
 b₁ = 3
 c₁ = - (p-3)


12x + py - p=0

a₂ = 12
b₂ = p
c₂ = -p

The equations have infinite solutions

a₁/a₂ = b₁/b₂ => c₁/c₂

p/12 = 3/p =>  p-3/p


p/12 => 3/p

We need to cross multiply,

p² =>36

p => √36

p = 6

 p = 6 is the correct answer


Hope it helps you!