For which values of $ a $ and $ b $, will the following pair of linear equations have infinitely many solutions?
$ x+2 y=1 $
$ (a-b) x+(a+b) y=a+b-2 $


Given: 

Equations: $x+2y=1$; $(a−b)x+(a+b)y=2$

To do: 

We have to find the values of $a$ and $b$, for which the following pair of linear equations will have infinitely many solutions.

Solution:

The condition for infinitely many solutions is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

From ratio I and II

$2a-2b=a+b$

$\Rightarrow a-3b=0\ ....( i)$

From ratio II and III

$2a+2b-4=a+b$

$\Rightarrow a+b=4\ .....( ii)$ 

Now solving $( i)$ and $( ii)$ we have

$a-3b=0\ ......( i)$

$a+b=4\ ......( ii)$  [Subtracting $( ii)$ from $( i)$ ]

$-4b=-4$

$\Rightarrow b=1$

and $a=4-b$

$\Rightarrow a=4-1$         [from $( ii)$]

$\Rightarrow a=3$

The values of \( a \) and \( b \) are $3$ and $1$ respectively.

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Updated on: 10-Oct-2022

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