Question

Find the values of $pandq$ for which the following system of linear equations has infinite number of solution .

$2x+3y=7;(p+q)x+(2p-q)y=3(p+q-1)$

Answer 1

Step 1:If equations has infinite number of solutions

then,

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

Given,

$$\begin{gathered} 2x+3y-7=0....\left(1\right) \\ (p+q)x+(2p-q)y=3(p+q-1)=0...\left(2\right) \end{gathered}$$

Equation (1) and (2) are in standard form

$$\begin{gathered} \text{here}{a}_1=2b_2=3c_1=-7 \\ {a}_2=p+q{b}_2=2p-q{c}_2=-3(p+q-1) \\ i.e\frac{2}{p+q}=\frac{3}{2p-q}=\frac{-7}{-3(p+q-1)} \end{gathered}$$

Step 2:Solving the equation by cross multiplication rule

$$\begin{gathered} \text{By taking}\frac{2}{p+q}=\frac{3}{2p-q} \\ \Rightarrow 2(2p-q)=3(p+q) \\ \Rightarrow 4p-2q=3p+3q \\ \Rightarrow 4p-2q-3p-3q=0 \\ \Rightarrow p-5q=0....\left(3\right) \\ \text{Taking}\frac{2}{p+q}=\frac{7}{3(p+q-1)} \\ \Rightarrow 6p+6q-6=7p+7q \\ \Rightarrow 7p+7q-6p-6q=-6 \\ \Rightarrow p+q=-6....\left(4\right) \end{gathered}$$

Step 3:Solving the equations to find $pandq$

Solving $\left(3\right)$ and $\left(4\right)$

$$\begin{gathered} \left(3\right)-\left(4\right)\Rightarrow -5q-q=0-(-6) \\ \Rightarrow -6q=6 \\ \Rightarrow q=-1 \end{gathered}$$

Substituting $q=-1\text{in}\left(3\right)$

$$\begin{gathered} \Rightarrow p=5q \\ \Rightarrow p=-5 \end{gathered}$$

Therefore the value of $p=-5andq=-1$