Question

Find the value of k for which the system of linear equations has an infinite number of solutions:
kx + 3y = (2k + 1),
2(k + 1)x + 9y = (7k + 1).

Answer 1

The given system of equations:
kx + 3y = (2k + 1)
⇒ kx + 3y - (2k + 1) = 0                   ….(i)

And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y - (7k + 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 = k, b_1= 3, c_1= -(2k + 1) and a_2 = 2(k + 1), b_2 = 9, c_2= -(7k + 1)`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
` i.e., k/(2(k+1)) = 3/9 = (−(2k+1))/(−(7k+1))`
`⇒k/(2(k+1)) = 1/3 = ((2k+1))/((7k+1))`
Now, we have the following three cases:
Case I:
`k/(2(k+1)) = 1/3`
⇒ 2(k + 1) = 3k
⇒ 2k + 2 = 3k
⇒ k = 2
Case II:
`1/3 = ((2k+1))/((7k+1))`
⇒ (7k + 1) = 6k + 3
⇒ k = 2
Case III:
`k/(2(k+1)) = ((2k+1))/((7k+1))`
⇒ k(7k + 1) = (2k + 1) × 2(k + 1)
`⇒ 7k^2 + k = (2k + 1) (2k + 2)`
`⇒ 7k^2 + k = 4k^2 + 4k + 2k + 2`
`⇒ 3k^2 – 5k – 2 = 0`
`⇒ 3k^2 – 6k + k – 2 = 0`
⇒ 3k(k – 2) + 1(k – 2) = 0
⇒ (3k + 1) (k – 2) = 0
`⇒ k = 2 or k = (−1)/3`
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.