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

Given: 

The given system of equations is:

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

To do: 

We have to determine 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:

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

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 $a_2=(a-1), b_2=a+2, c_2=-3a$

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}} \ $

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

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

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

$6a=7a-7$

$7a-6a=7$

$a=7$

The value of $a$ for which the given system of equations has infinitely many solutions is $7$.

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Updated on: 10-Oct-2022

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