Question

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

• A. $6$
• B. $-6$
• C. $0$
• D. $2$

Answer 1

The correct option is A $6$

px+3y−(p−3)=0px+3y−(p−3)=0 ........ (I)
12x+py−p=012x+py−p=0 ........(II)
For infinitely many solution
$\frac{p}{12}=\frac{3}{p}=\frac{-(p-3)}{(-p)}$
From I and II
$p^2=36\Rightarrow p^2-36=0\Rightarrow p=\pm 6$ p=0

From II and III
$3=p-3\Rightarrow p=6$
So, $p=6$