Question

Find the values of a and b for which the following system of equations has infinitely many solutions:

(a - 1)x + 3y = 2

6x + (1 + 2b)y = 6

Answer 1

The given system of equations is

(a - 1)x + 3y - 2 = 0

6x + (1 + 2b)y - 6 = 0

It is of the form

`a_1x + b_1y + c_1 = 0` `

a_2x + b_2y + c_2 = 0`

Where `a_1 = a - 1, b_1 = 3, c_1 = -2`

And `a_2 = 6, b_2 = 1 -2b, c_2 = -6`

The given system of equations will be have infinite number of solutions, if

`a_1/a_2 = b_1/b_2 = c_1/c_2`

`=> (a - 1)/6 = 3/(1 - 2b) = (-2)/(-6)`

`=> (a - 1)/6 = 3/(1 - 2b) = 1/3`

`=> (a- 1)/b = 1/3 and 3/(1 - 2b) = 1/3`

`=> 3(a - 1) = 6 and 3 xx 3 = 1 - 2b`

`=> a - 1 = 2 and 9 = 1 - 2b`

`=> a = 2 + 1 and 2b = 1 - 9`

`=> a = 3 and 2b = -8`

`=> a = 3 and b = (-8)/2 = -4`

Hence, the given system of equations will have infinitely many solutions,

if a = 3 and b = -4