Determine the values of $ a $ and $ b $ so that the following system of linear equations have infinitely many solutions:
$(2 a-1) x+3 y-5=0$
$3 x+(b-1) y-2=0$

Given: 

The given system of equations is:

$(2 a-1) x+3 y-5=0$
$3 x+(b-1) y-2=0$

To do: 

We have to determine the values of $a$ and $b$ so that the given system of equations has infinitely many solutions.

Solution:

The given system of equations is:

$(2 a-1) x+3 y-5=0$
$3 x+(b-1) y-2=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

Comparing the given system of equations with the standard form of equations, we have,

$a_1=(2a-1), b_1=3, c_1=-5$ and $a_2=3, b_2=(b-1), c_2=-2$

The condition for which the given system of equations has infinitely many solutions is

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} \ $

$\frac{2a-1}{3}=\frac{3}{b-1}=\frac{-5}{-2}$

$\frac{2a-1}{3}=\frac{3}{b-1}=\frac{5}{2}$

$\frac{2a-1}{3}=\frac{5}{2}$ and $\frac{3}{b-1}=\frac{5}{2}$

$(2a-1)\times2=5\times3$ and $3\times2=5\times(b-1)$

$4a-2=15$ and $6=5b-5$

$4a=15+2$ and $5b=6+5$

$4a=17$ and $5b=11$

$a=\frac{17}{4}$ and $b=\frac{11}{5}$

The values of $a$ and $b$ for which the given system of equations has infinitely many solutions is $\frac{17}{4}$ and $\frac{11}{5}$ respectively. 

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Updated on: 10-Oct-2022

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