Question

For what value of p will the following pair of linear equations have
infinitely many solutions (p-3)x+3y=6 ; px+py=12

Answer 1

Hi there !!

For infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

p-3/p = 3/p = 6/12

p= 6

Thankyou :)

Answer 2

Linear Equations in 2 Variables :

Solution:

Equation 1 : ( p - 3)x + 3y = 6

( p - 3)x + 3y - 6 = 0

Equation 2 : px + py - 12 = 0

Now, For Infinitely many solutions or parallel consistency of equations ,

$</strong><strong>\</strong><strong>boxed</strong><strong>{</strong><strong>\mathsf{\dfrac{a_1}{a_2}\:=\:{\dfrac{b_1}{b_2}\:=\:{\dfrac{c_1}{c_2}</strong><strong>}</strong><strong>}</strong><strong>}</strong><strong>}</strong><strong>$

Here, $\mathsf{a_1\:=\:(p\:-\:3)}$, $\mathsf{a_2\:=\:p}$ , $\mathsf{b_1\:=\:3}$, $\mathsf{b_2\:=\:p}$ , $\mathsf{c_1\:=\:-6}$ , $\mathsf{c_2\:=\:-12)}$

Now, putting the values,

$\mathsf{\dfrac{(p\:-\:3)}{p}\:=\:{\dfrac{3}{p}\:=\:{\dfrac{-6}{-12}}}}$

$\mathsf{\dfrac{(p\:-\:3)}{p}\:=\:{\dfrac{-6}{-12}}}$

$\mathsf{\dfrac{(p\:-\:3)}{p}\:=\:{\dfrac{1}{2}}}$

2( p - 3) = p

2p - 6 = p

2p - p = 6

p = 6

Also,

$\mathsf{\dfrac{(p\:-\:3)}{p}\:=\:{\dfrac{3}{p}}}$

p ( p - 3) = 3p

$\mathsf{p^2\:-\:3p\:=\:3p}$

$\mathsf{p^2\:=\:6p}$

p = 6.

⛛ $\mathsf{\green{Value\:of\:p\:is\:6</u>}}$.

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