Question

For what values of $p$ will the following pair of linear equations have infinitely many solutions
$\begin{array}{c}\left(p-3\right)x+3y=p\\ px+py=12\end{array}$

Answer 1

Given that

$(p-3)x+3y=p....\left(1\right)$

$px+py=\mathrm{12....}\left(2\right)$

also given that

pair of the linear equation have infinitely solution then

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{p-3}{p}=\frac{3}{p}=\frac{p}{12}$

if,

$\Rightarrow \frac{3}{p}=\frac{p}{12}$

$\Rightarrow p^2=36$

$\Rightarrow p^2=6^2$

$\therefore p=6$

Hence this is the answer