Question

Find the values of $a$ and $b$ for which the following system of linear equations has infinite number of solutions?

$2x+3y=7;(a+b+1)x+(a+2b+2)y=4(a+b)+1$

Answer 1

Step 1: If the pair of linear equations has infinite number of solutions

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

Given,

$$\begin{gathered} 2x+3y-7=0..................................\left(1\right) \\ (a+b+1)x+(a+2b+2)y-\left[4\right(a+b)+1]=0...................\left(2\right) \end{gathered}$$

$$\begin{gathered} \text{here}{a}_1=2,b_1=3,c_1=-7 \\ {a}_2=a+b+1,b_2=a+2b+2,c_2=-\left[4\right(a+b)+1] \end{gathered}$$

Step 2:Substitution of values in the condition $\frac{{\mathbf{a}}_{\mathbf{1}}}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{b}}_{\mathbf{1}}}{{\mathbf{b}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{c}}_{\mathbf{1}}}{{\mathbf{c}}_{\mathbf{2}}}$ and solving by cross multiplication method

$$\begin{gathered} \Rightarrow \frac{2}{a+b+1}=\frac{3}{a+2b+2}=\frac{-7}{-\left[4\right(a+b)+1]} \\ \text{Taking},\frac{2}{a+b+1}=\frac{3}{a+2b+2} \\ \Rightarrow 2(a+2b+2)=3(a+b+1) \\ \Rightarrow 2a+4b+4=3a+3b+3 \\ \Rightarrow 2a+4b+4-3a-3b-3=0 \\ \Rightarrow -a+b+1=0 \\ \Rightarrow a-b=1.................................\left(3\right) \end{gathered}$$

$$\begin{gathered} \text{Taking},\frac{2}{a+b+1}=\frac{7}{4a+4b+1} \\ \Rightarrow 2[4a+4b+1]=7[a+b+1] \\ \Rightarrow 8a+8b+2=7a+7b+7 \\ \Rightarrow 8a+8b+2-7a-7b-7=0 \\ \Rightarrow a+b-5=0 \\ \Rightarrow a+b=5................................\left(4\right) \end{gathered}$$

Step 3:Solving the equations $\left(3\right)$ and $\left(4\right)$ to get the values of $a$ and $b$

$$\begin{gathered} \left(3\right)+\left(4\right)\Rightarrow 2a=6 \\ \Rightarrow a=3 \end{gathered}$$

Using equation$\left(4\right)$

$$\begin{gathered} \left(4\right)\Rightarrow a+b=5 \\ \Rightarrow b=5-a \\ \Rightarrow b=5-3 \\ \Rightarrow b=2 \end{gathered}$$

Hence the values of $a=3andb=2$