Question

Question 4 (v)
Find the values of p and q for the following pair of equations 2x + 3y = 7 and 2px + py = 28 – qy, if the pair of equations has infinitely many solutions.

Answer 1

Given system of linear equations is
$2px+py=28-qy$
i.e., $2px+(p+q)y=28$ ...(i)
$2x+3y=7$ ...(ii)

Comparing with $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, we get

$a_1=2p,b_1=p+qand{c}_1=-28$

$a_2=2,b_2=3and{c}_2=-7$

The system of equations will have infinitely many solutions if

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \underset{I}{\frac{2p}{2}}=\underset{II}{\frac{p+q}{3}}=\underset{III}{\frac{28}{7}}$
Using ratios I and II we get,
$\frac{2p}{2}=\frac{p+q}{3}$
$\frac{p}{1}=\frac{p+q}{3}$
$\Rightarrow 3p=p+q$
$\Rightarrow 2p-q=0$
$\Rightarrow q=2p$ ...(iii)
Using ratios I and III, we get
$\frac{2p}{2}=\frac{28}{7}\Rightarrow p=4$
$\therefore q=2p=2\times 4=8$ [From (iii)]
$\therefore$ q = 8, and p =4
Now, $\frac{2p}{2}=\frac{p+q}{3}=\frac{28}{7}$
$\Rightarrow \frac{p}{1}=\frac{p+q}{3}=\frac{4}{1}$
By substituting the values of p and q, we have
$4=\frac{4+8}{3}=4$
$\Rightarrow 4=4=4$
Hence, the given system of equations has infinitely many solutions when p = 4 and q = 8.