Question

For what value of p, the following pair of linear equation in two variables will have infinitely many solutions? Px +3y-(p-3) =0 12x +py - p = 0

Answer 1

use condition for infinitely many solutions,
e.g., $\boxed{\bf{\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}}}$

given, px + 3y - (p - 3) = 0
12x + py - p = 0

here, $a_1=p,b_1=3,c_1=-(p-3)$
$a_2=12,b_2=p,c_2=-p$

so, p/12 = 3/p = -(p - 3)/-p

p/12 = 3/p

p² = 12 × 3 = 36 = 6²

p = ± 6 .....(1)

again, 3/p = (p-3)/p

3p = p² - 3p

6p = p²

p = 0, 6 .......(2)

from equations (1) and (2),
common value of p = 6
hence, value of p = 6

Answer 2

Given:

Linear equation 1 =  Px + 3y - ( p - 3 ) = 0

Linear equation 2 = 12x +py - p = 0

To find:

The value of p, the following pair of linear equation in two variables will have infinitely many solutions.

Solution:

Condition for infinitely many solutions,

a1 / a2 = b1 / b2 = c1 / c2

Here,

a1= p

b1 = 3

c1 = - ( p - 3 )

a2 = 12

b2 = p

c2 = -p

Substituing,

p / 12 = 3 / p = - ( p - 3 ) / -p  

p / 12 = 3 / p

Solving,

p^2 = 36

Hence,

p = ± 6

If p = -6, the condition cannot be satisfied.

Hence,

p = 6

Hence, the pair of linear equation in two variables will have infinitely many solutions if p = 6.