Question

Find the value of k for which each of the following system of equations has infinitely many solutions :

2x +3y = k

(k - 1)x + (k + 2)y = 3k

Answer 1

The given system of the equation may be written as

2x +3y = k = 0

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

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

where `a_1 = 2, b_1 = 3, c_1= -k`

And `a_2 = k -1,b_2 = k + 2, c_2 = 3k`

For a unique solution, we must have

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

`=> 2/(k-1) = 3/(k +1) = (-k)/(-3k)`

`=> 2/(k -1) = 3/(k +1) and 3/(k +1) = (-k)/(-3k)`

`=> 2(k + 2) = 3(k - 1) and 3 xx 3 = k + 2`

`=> 2k + 4 = 3k - 3 and  9 = k + 2`

`=> 4 + 3 = 3k - 2k  and 9 - 2 = k`

=> 7 = k and 7 = k

 k = 7 satisfies both the conditions

Hence, the given system of equations will have infinitely many solutions if  k = 7