Question

Find the values of a and b for which the system of linear equations has an infinite number of solutions:
2x + 3y = 7, (a + b + 1)x - (a + 2b + 2)y = 4(a + b) + 1.

Answer 1

The given system of equations can be written as
2x + 3y = 7
⇒2x + 3y - 7 = 0 ….(i)
and (a + b + 1)x - (a + 2b + 2)y = 4(a + b) + 1
(a + b + 1)x - (a + 2b + 2)y – [4(a + b) + 1] = 0 ….(ii)
These equations are of the following form:
${\displaystyle a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0}$
where, ${\displaystyle a_1=2,b_1=3,c_1=-7\text{and}{a}_2=(a+b+1),b_2=(a+2b+2),c_2=–[4(a+b)+1]}$
For an infinite number of solutions, we must have:
${\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}}$
${\displaystyle \frac{2}{(a+b+1)}=\frac{3}{(a+2b+2)}=\frac{-7}{-[4(a+b)+1]}}$
${\displaystyle \Rightarrow \frac{2}{(a+b+1)}=\frac{3}{(a+2b+2)}=\frac{7}{[4(a+b)+1]}}$
${\displaystyle \Rightarrow \frac{2}{(a+b+1)}=\frac{3}{(a+2b+2)}\text{and}\frac{3}{(a+2b+2)}=\frac{7}{[4(a+b)+1]}}$
⇒ 2(a + 2b + 2) = 3(a + b + 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
⇒ a – b - 1=0 and 12a + 12b + 3 = 7a + 14b + 14
⇒ a - b = 1 and 5a – 2b = 11
a = (b + 1)                  …….(iii)
5a - 2b = 11                  ……(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) – 2b = 11
⇒5b + 5 – 2b = 11
⇒ 3b = 6
⇒ b = 2
On substituting b = 2 in (iii), we get:
a = 3

∴a = 3 and b = 2.