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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.