Question

Determine the values of $mandn$ so that the following linear equations have infinite solutions$(2m-1)x+3y-5=0;3x+(n-1)y-2=0$

Answer 1

Step 1: By writing the equations in the standard form

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

Step 2: If the linear equations have infinite solutions

$$\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ Here, \\ {a}_1=(2m-1)b_1=3c_1=-5 \\ {a}_2=3b_2=(n-1)c_2=-2 \\ i.e,\frac{2m-1}{3}=\frac{3}{n-1}=\frac{-5}{-2} \\ \Rightarrow \frac{2m-1}{3}=\frac{3}{n-1}=\frac{5}{2}........................\left(3\right) \end{gathered}$$

Step 3: By applying the cross multiplication method

$$\begin{gathered} From\left(3\right), \\ \frac{2m-1}{3}=\frac{5}{2} \\ \Rightarrow 2(2m-1)=5x3 \\ \Rightarrow 4m-2=15 \\ \Rightarrow 4m=15+2 \\ \Rightarrow 4m=17 \\ \Rightarrow m=\frac{17}{4} \\ From\left(3\right)taking, \\ \frac{3}{n-1}=\frac{5}{2} \\ \Rightarrow 3x2=5(n-1) \\ \Rightarrow 6=5n-5 \\ \Rightarrow 6+5=5n \\ \Rightarrow 5n=11 \\ \Rightarrow n=\frac{11}{5} \end{gathered}$$

Step 4: For checking, substituting the values of $mandn$ in equation (3)

$$\begin{gathered} \frac{2m-1}{3}=\frac{2x{\displaystyle \frac{17}{4}}-1}{3} \\ =\frac{{\displaystyle \frac{17}{2}}-1}{3} \\ =\frac{17-2}{2x3} \\ =\frac{15}{2x3} \\ =\frac{5}{2} \\ \frac{3}{n-1}=\frac{3}{{\displaystyle \frac{11}{5}}-1} \\ =\frac{3}{{\displaystyle \frac{11-5}{5}}} \\ =\frac{3x5}{6} \\ =\frac{5}{2} \end{gathered}$$

Therefore the values of $m=\frac{17}{4}andn=\frac{11}{5}.$