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 equations are $(p-3)x+3y=p$ and $px+py=12$

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

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

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

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

$\Rightarrow 12p-36=p^2$

$\Rightarrow p^2-12p=-36$

$\Rightarrow p^2-12p+36$

$\Rightarrow p^2-6p-6p+36=0$

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

$\Rightarrow (p-6)(p-6)$

$\Rightarrow p-6=0$

$\therefore p=6$