Question

Question 3
For what values of a and b will the following pair of linear equations have infinitely many solutions?
x + 2y = 1
(a - b)x + (a + b)y = a + b - 2

Answer 1

Given pair of linear equations are
x + 2y = 1
and (a – b) x + (a + b) y = a + b – 2
on comparing with ax + by +c = 0 , we get
$a_1=1,b_1=2and{c}_1=-1a_2=(a–b),b_2=(a+b)and{c}_2=-(a+b–2)$
For infinitely many solutions of the pair of linear equations
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\Rightarrow \frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{(a+b-2)}$
Taking first two parts
$\frac{1}{a-b}=\frac{2}{a+b}\Rightarrow a+b=2a-2b\Rightarrow 2a-a=2b+b\Rightarrow a=3b...\left(i\right)$
Taking last two parts
$\frac{2}{a+b}=\frac{1}{(a+b-2)}\Rightarrow 2a+2b-4=a+b\Rightarrow a+b=4...\left(ii\right)$
Now, put the value of 'a' from Eq (i) in Eq. (ii), we get
3b + b = 4
$\Rightarrow$ 4b = 4
$\Rightarrow$ b = 1
Put the value of b in Eq. (i), we get
a = 3 $\times$ 1
$\Rightarrow$ a = 3
So, the values (a,b) = (3,1) satisfies the given condition. Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.