Question

Find the value of $k$ for which the following system of equations has infinitely many solutions

$(k-1)x+3y=7;(k+1)x+6y=(5k-1)$$=3$

Answer 1

Step 1: By writing the equations in the standard form

$$\begin{gathered} (k-1)x+3y-7=0.................................\left(1\right) \\ (k+1)x+6y-(5k-1)=0.........................\left(2\right) \end{gathered}$$

Step 2: If a pair of linear equations have infinitely many solutions

$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ here, \\ {a}_1=k-1b_1=3c_1=-7 \\ {a}_2=k+1b_2=6c_2=-(5k-1) \\ i.e, \\ \frac{k-1}{k+1}=\frac{3}{6}=\frac{-7}{-(5k-1)} \\ \Rightarrow \frac{k-1}{k+1}=\frac{3}{6}=\frac{7}{5k-1}...........................\left(3\right) \end{gathered}$$

Step 3: By applying the cross multiplication method

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

Step 4: For checking, substituting the value of $k$ in equation(3)

$$\begin{gathered} \frac{k-1}{k+1}=\frac{3-1}{3+1} \\ =\frac{2}{4} \\ =\frac{1}{2} \\ \frac{7}{5k-1}=\frac{7}{14} \\ =\frac{1}{2} \end{gathered}$$

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

Therefore the value of $k$ $=3.$