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)x - (a + b – 3)y = 4a + b.

Answer 1

The given system of equations can be written as
2x - 3y = 7
⇒2x - 3y - 7 = 0                        ….(i)
and (a + b)x - (a + b – 3)y = 4a + b
⇒(a + b)x - (a + b – 3)y - 4a + b = 0               ….(ii)
These equations are of the following form:
`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`
Here,` a_1 = 2, b_1= -3, c_1= -7 and a_2 = (a + b), b_2= -(a + b - 3), c_2= -(4a + b)`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`2/(a+b) = (−3)/(−(a+b−3)) = (−7)/(−(4a+b))`
`⇒ 2/(a+b) = 3/((a+b−3)) = 7/((4a+b))`
`⇒ 2/(a+b) = 7/((4a+b)) and 3/((a+b−3)) = 7/((4a+b))`
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b - 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b - 21
⇒ 4a = 17 and 5b = 11
∴ a = 5b                              …….(iii)
and 5a = 4b – 21               ……(iv)
On substituting a = 5b in (iv), we get:

25b = 4b – 21
⇒21b = -21
⇒b = -1
On substituting b = -1 in (iii), we get:
a = 5(-1) = -5
∴a = -5 and b = -1.