Question

Find the value of k for which the system of linear equations has an infinite number of solutions:
2x + (k – 2)y = k,
6x + (2k - 1)y = (2k + 5).

Answer 1

The given system of equations:

2x + (k – 2)y = k
⇒ 2x + (k – 2)y - k = 0                ….(i)
And, 6x + (2k - 1) y = (2k + 5)
⇒ 6x + (2k - 1) y - (2k + 5) = 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 = 2, b_1= (k – 2), c_1= -k and a_2 = 6, b_2 = (2k - 1), c_2= -(2k + 5)`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`2/6 = ((k−2))/((2k−1)) =( −k)/(−(2k+5))`
`⇒1/3 = ((k−2))/((2k−1)) = k/((2k+5))`
Now, we have the following three cases:
Case I:
`1/3 = ((k−2))/((2k−1))`
⇒ (2k – 1) = 3(k – 2)
⇒ 2k - 1 = 3k – 6 ⇒ k = 5
Case II:
`((k−2))/((2k−1)) = k/((2k+5))`
⇒ (k - 2) (2k + 5) = k(2k - 1)
`⇒ 2k^2 + 5k – 4k – 10 = 2k^2 – k`
⇒ k + k = 10 ⇒ 2k = 10 ⇒ k = 5
Case III:
`1/3 = k/((2k+5))`
⇒ 2k + 5 = 3k ⇒ k = 5
Hence, the given system of equations has an infinite number of solutions when k is equal to 5.