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

Given: 

The given system of equations is:

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

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:

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

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,

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

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}} \ $

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

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

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

$2\times(a+b)=3\times(a-b)$ and $3\times(3a+b-2)=7\times(a+b)$

$2a+2b=3a-3b$ and $9a+3b-6=7a+7b$

$3a-2a=2b+3b$ and $9a-7a+3b-7b=6$

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

Using $a=5b$ in $2a-4b=6$, we get,

$2(5b)-4b=6$

$10b-4b=6$

$6b=6$

$b=1$

This implies,

$a=5(1)=5$

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

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Updated on: 10-Oct-2022

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