Find the values of $ a $ and $ b $ for which the following system of equations has infinitely many solutions:
$(a-1)x+3y=2$
$6x+(1-2b)y=6$

Given: 

The given system of equations is:

$(a-1)x+3y=2$
$6x+(1-2b)y=6$

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 can be written as:

$(a-1)x+3y-2=0$
$6x+(1-2b)y-6=0$

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,

$a_1=(a-1), b_1=3, c_1=-2$ and $a_2=6, b_2=(1-2b), c_2=-6$

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}} \ $

$\frac{(a-1)}{6}=\frac{3}{1-2b}=\frac{-2}{-6}$

$\frac{a-1}{6}=\frac{3}{1-2b}=\frac{1}{3}$

$\frac{a-1}{6}=\frac{1}{3}$ and $\frac{3}{1-2b}=\frac{1}{3}$

$3\times(a-1)=1\times6$ and $3\times3=1\times(1-2b)$

$3a-3=6$ and $9=1-2b$

$3a=6+3$ and $2b=1-9$

$3a=9$ and $2b=-8$

$a=\frac{9}{3}$ and $b=\frac{-8}{2}$

$a=3$ and $b=-4$

The values of $a$ and $b$ for which the given system of equations has infinitely many solutions is $3$ and $-4$ respectively.   

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Updated on: 10-Oct-2022

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