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:
`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`
where, `a_1 = 2, b_1= 3, c_1= -7 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:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`2/((a+b+1)) = 3/((a+2b+2)) = (−7)/(−[4(a+b)+1])`
`⇒ 2/((a+b+1)) = 3/((a+2b+2)) = 7/([4(a+b)+1])`
`⇒ 2/((a+b+1)) = 3/((a+2b+2))and 3/((a+2b+2)) = 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.