Question

Find the value of k for which each of the following systems of equations have infinitely many solution:
2x+3y=7(k+1)x+(2k-1)y=4k+1

Answer 1

The value of k for which each of the following systems of equations have infinitely many solution:

2x + 3y = 7(k+1)x + (2k-1)y = 4k+1

• It can be written as,

2x + 3y - (4k + 1) = 7(k+1)x + (2k-1)y - (4k+1)

a1 = 2 , b1 = 3, c2 = -(4k + 1) and

a2 = 7(k+1), b2 = (2k-1), c2= -(4k + 1)

• Condition for infinitely many solutions is

a1/a2 = b1/b2 = c1/c2

°•° 2/7(k+1) = 3/(2k-1) = -(4k + 1)/-(4k + 1)

2/7(k+1) = 1

2 = 7k + 7

7k = -5 => k = -5/7

Or

3/2k-1 = 1 => 3 = 2k - 1

2k = 4 => k = 2