Find the values of $ a $ and $ b $ for which the following system of equations has infinitely many solutions:
$2x+3y-7=0$
$(a-1)x+(a+1)y=(3a-1)$

Given: 

The given system of equations is:

$2x+3y-7=0$
$(a-1)x+(a+1)y=(3a-1)$

To do: 

We have to find the value of $a$ so that the given system of equations has infinitely many solutions.

Solution:

The given system of equations can be written as:

$2x+3y-7=0$

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

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,

$a_1=2, b_1=3, c_1=-7$ and $a_2=(a-1), b_2=a+1, c_2=-(3a-1)$

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}} \ $

$\frac{2}{a-1}=\frac{3}{a+1}=\frac{-7}{-(3a-1)}$

$\frac{2}{a-1}=\frac{7}{3a-1}$

$2\times(3a-1)=7\times(a-1)$

$6a-2=7a-7$

$7a-6a=7-2$

$a=5$

The value of $a$ for which the given system of equations has infinitely many solutions is $5$. 

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Updated on: 10-Oct-2022

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