Question

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

Answer 1

Given equation are

$(p-3)x+3y=p$

$px+py=12$

The pair of linear equation has infinitely many solutions.

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

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

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

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

$(p-3)12=p^2$

$12p-36=p^2$

$p^2-12p-36=0$

$p^2-6p-6p-36=0$

$p(p-6)-6(p-6)=0$

$p-6=0$

$\therefore p=6$.