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

Given: 

The given system of equations is:

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

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:

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

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,

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

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}} \ $

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

$\frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{a+b-2}$

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

$1\times(a+b)=2\times(a-b)$ and $1\times(a+b-2)=1\times(a-b)$

$a+b=2a-2b$ and $a+b-2=a-b$

$2a-a=b+2b$ and $a-a+b+b=2$

$a=3b$ and $2b=2$

$a=3b$ and $b=1$

$a=3(1)=3$ and $b=1$

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

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

24 Views

Kickstart Your Career

Get certified by completing the course

Get Started