Question

For what value of $k$ will the following system of linear equations has infinite number of solutions

$2x+3y=2;(k+2)x+(2k+1)y=2(k-1)$

Answer 1

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

Applying condition,

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

Given,

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

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

$$\begin{gathered} here{a}_1=2b_1=3c_{1=}-2 \\ {a}_{2=}k+2b_2=2k+1_{}{c}_2=-2(k-1) \\ Here,\frac{2}{k+2}=\frac{3}{2k+1}=\frac{-2}{-2(k-1)} \\ \Rightarrow \frac{2}{k+2}=\frac{3}{2k+1}=\frac{1}{k-1} \end{gathered}$$

Step 2:Find the value of $\mathit{k}$by cross multiplication method

$$\begin{gathered} taking\frac{2}{k+2}=\frac{3}{2k+1} \\ \Rightarrow 2(2k+1)=3(k+2) \\ \Rightarrow 4k+2=3k+6 \\ \Rightarrow 4k-3k=6-2 \\ \Rightarrow k=4 \end{gathered}$$

Checking,

$$\begin{gathered} \frac{2}{k+2}=\frac{2}{4+2}=\frac{2}{6}=\frac{1}{3} \\ \frac{3}{2k+1}=\frac{3}{8+1}=\frac{3}{9}=\frac{1}{3} \\ \frac{1}{k-1}=\frac{1}{4-1}=\frac{1}{3} \end{gathered}$$

Therefore,$\frac{2}{k+2}=\frac{3}{2k+1}=\frac{1}{k-1}$

Hence the value of $k=4$