Question

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

Answer 1

This system of equation is of the form
$a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$

where $a_1=2m-1,b_1=3,c_1=-5$

and $a_2=3,b_2=n-1$ and $c_2=-2$

For infinitely many solutions, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

The given system of equations will have infinite number of solutions, if

$\frac{2m-1}{3}=\frac{3}{n-1}=\frac{-5}{-2}$

$\Rightarrow \frac{2m-1}{3}=\frac{3}{n-1}=\frac{5}{2}$

$\Rightarrow \frac{2m-1}{3}=\frac{5}{2}$ and $\frac{3}{n-1}=\frac{5}{2}$

$\Rightarrow 4m-2=15$ and $6=5n-5$

$\Rightarrow 4m=17$ and $5n=11$

$\Rightarrow m=\frac{17}{4}$ and $n=\frac{11}{5}$

Hence, the given system of equations will have infinite number of solutions, if $m=\frac{17}{4}$ and $n=\frac{11}{5}$.