Find the values of $ a $ and $ b $ for which the following system of equations has infinitely many solutions:
$(2 a-1) x-3 y=5$
$3 x+(b-2) y=3$

Given: 

The given system of equations is:

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

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:

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

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 $a_2=3, b_2=(b-2), c_2=-3$

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}} \ $

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

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

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

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

$6a-3=15$ and $-9=5b-10$

$6a=15+3$ and $5b=10-9$

$6a=18$ and $5b=1$

$a=\frac{18}{6}=3$ and $b=\frac{1}{5}$

The values of $a$ and $b$ for which the given system of equations has infinitely many solutions is $3$ and $\frac{1}{5}$ respectively. 

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Updated on: 10-Oct-2022

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