Question

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

$x+(k+1)y=5;(k+1)x+9y=8k-1$

Answer 1

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

Applying condition,

$$\begin{gathered} x+(k+1)y=5 \\ \Rightarrow x+(k+1)y-5=0...........................................\left(1\right) \\ (k+1)x+9y=8k-1 \\ \Rightarrow (k+1)x+9y-(8k-1)=0...................................\left(2\right) \end{gathered}$$

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

$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ here{a}_1=1b_1=k+1c_1=-5 \\ {a}_2=k+1b_2=9c_2=-(8k-1) \\ here\frac{1}{k+1}=\frac{k+1}{9}=\frac{-5}{-(8k-1)} \\ \Rightarrow \frac{1}{k+1}=\frac{k+1}{9}=\frac{5}{8k-1} \end{gathered}$$

Step 2:To find the value of $k$ using cross multiplication method

$$\begin{gathered} Taking\frac{1}{k+1}=\frac{k+1}{9} \\ \Rightarrow {(k+1)}^2=9takingsquarerootonbothsides \\ i.ek+1=\pm 3 \\ k+1=3 \\ \Rightarrow k+2 \\ k+1=-3 \\ \Rightarrow k=-4 \end{gathered}$$

Step 3: Verification when $k=2andk=-4$

$$\begin{gathered} whenk=2, \\ \Rightarrow \frac{1}{k+1}=\frac{1}{2+1}=\frac{1}{3} \\ \Rightarrow \frac{k+1}{9}=\frac{2+1}{9}=\frac{3}{9}=\frac{1}{3} \\ \Rightarrow \frac{5}{8k-1}=\frac{5}{16-1}=\frac{5}{15}=\frac{1}{3} \\ here\frac{1}{k+1}=\frac{k+1}{9}=\frac{5}{8k-1} \\ whenk=-4, \\ \Rightarrow \frac{1}{k+1}=\frac{1}{-4+1}=\frac{1}{-3}=\frac{-1}{3} \\ \Rightarrow \frac{k+1}{9}=\frac{-4+1}{9}=\frac{-3}{9}=\frac{-1}{3} \\ \frac{5}{8k-1}=\frac{5}{8(-4)-1}=\frac{5}{-32-1}=\frac{5}{-33}=\frac{-5}{33} \\ here\frac{1}{k+1}=\frac{k+1}{9}\ne \frac{5}{8k-1} \end{gathered}$$

Therefore the value of $k=2$