Question

for what value of "p" will the pair of equation px + 3y = p-3, 12x+py=p has no solution plz tell me correct answer​

Answer 1

The given equations are:

px + 3y = p - 3 and 12x + py = p

These equations are of the form:

$a_{1} x+b_{1} y=c_{1}$ and $a_{2} x+b_{2} y=c_{2}$

Here,  $a_{1}=p,b_{1} =3,c_{1}=p-3$ and  $a_{2}=12,b_{2} =p,c_{2}=p$

We have to find, the value of p.

Solution:

We know that:

The given system has no solution, we must have:

$\dfrac{a_{1} }{a_{2} } =\dfrac{b_{1} }{b_{2} }$ ≠ $\dfrac{c_{1} }{c_{2} }$

$\dfrac{p}{12 } =\dfrac{3 }{p }$ ≠ $\dfrac{p-3}{p}$

∴ $\dfrac{p}{12 } =\dfrac{3 }{p }$ ⇒ $p^{2}$ = 36 ⇒p = ± 6

Case 1: When p = 6, we get:

$\dfrac{6}{12 } =\dfrac{3 }{6 }=\dfrac{1}{2}$ and  $\dfrac{6-3}{6}=\dfrac{1}{2}$

$\dfrac{p}{12 } =\dfrac{3 }{p }$ ≠ is not satisfied.

Case 2: When p = - 6, we get:

$\dfrac{-6}{12 } =\dfrac{3 }{-6 }=-\dfrac{1}{2}$ and  $\dfrac{-6-3}{-6}=\dfrac{3}{2}$

$\dfrac{p}{12 } =\dfrac{3 }{p }$ ≠ $\dfrac{p-3}{p}$is not satisfied.

Thus, the required value of "p is - 6".

Answer 2

Given :  px + 3y = p-3,  12x+py=p

To Find : value of "p" the pair of equation  has no solution

Solution:

a₁x+ b₁y=c₁

a₂x+b₂y=c₂

Consistent

if a₁/a₂ ≠ b₁/b₂  ( unique solution)

a₁/a₂ = b₁/b₂ = c₁/c₂  ( infinite solution )

Inconsistent

if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂  ( No solution)

px + 3y = p-3

12x+py = p

no solution  if

p/12  = 3/p  ≠ (p - 3)/p

p/12  = 3/p

=> p² = 36

=> p  = ±6

P = 6  => p/12  = 3/p  = 1/2

(p - 3)/p = (6 - 3)/6 = 1/2

1/2 = 1/2

=> p/12  = 3/p  = (p - 3)/p  for p = 6

Hence Infinite solution   if p = 6

P = -6  => p/12  = 3/p  = -1/2

(p - 3)/p = (-6 - 3)/(-6) = 3/2

-1/2  ≠ 3/2

p/12  = 3/p  ≠ (p - 3)/p  if p = - 6

Hence no  solution   if p = -6

p = -6   , no solution

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