Question

Find the value of$k$, for which the following system of linear equations has infinite number of solutions

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

Answer 1

Step 1:If the pair of equation has infinite number of solutions

then,

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

Given,

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

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

Step 2:Substituting the values in the condition $\frac{{\mathbf{a}}_{\mathbf{1}}}{{\mathbf{a}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{b}}_{\mathbf{1}}}{{\mathbf{b}}_{\mathbf{2}}}\mathbf{}\mathbf{=}\mathbf{}\frac{{\mathbf{c}}_{\mathbf{1}}}{{\mathbf{c}}_{\mathbf{2}}}$

$$\begin{gathered} Here{a}_1=2b_1=3c_1=-7 \\ {a}_2=k-1b_2=k+2c_2=-3k \\ \Rightarrow \frac{2}{k-1}=\frac{3}{k+2}=\frac{-7}{-3k} \end{gathered}$$

Step 3: Solving the equation by cross multiplication method

$$\begin{gathered} Taking\frac{2}{k-1}=\frac{3}{k+2} \\ \Rightarrow 2(k+2)=3(k-1) \\ \Rightarrow 2k+4=3k-3 \\ \Rightarrow 3k-2k=4+3 \\ \Rightarrow k=7 \end{gathered}$$

Therefore the value of $k=7$