Question

Question 2 (i)
For which values of ‘a’ and ‘b’ does the following pair of linear equations have an infinite number of solutions?
2x + 3y =7
(a - b)x + (a + b)y = 3a +b - 2

Answer 1

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

Compare the given equations with

$a_{1}x+b_{1}y+c_1=0and{a}_{2}x+b_{2}y+c_2=0$.

$\frac{a_1}{a_2}=\frac{2}{a-b}$
$\frac{b_1}{b_2}=\frac{3}{a+b}$
$\frac{c_1}{c_2}=\frac{-7}{-(3a+b-2)}=\frac{7}{(3a+b-2)}$
For infinitely many solutions, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{2}{a-b}=\frac{7}{3a+b-2}$
$\Rightarrow$2(3a + b - 2) = 7 (a-b)
$\Rightarrow$6a + 2b - 4 = 7a - 7b
$\Rightarrow$a - 9b = -4 ... (i)
$\frac{2}{a-b}=\frac{3}{a+b}$
$\Rightarrow$2a + 2b = 3a - 3b
$\Rightarrow$ a - 5b = 0 ... (ii)
Subtracting equation (i) from (ii), we get
4b = 4
$\Rightarrow$ b = 1
Putting this value in equation (ii), we get
a - 5 $\times$ 1 = 0
$\Rightarrow$ a = 5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.