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Question
Find the value of k for which the system of linear equations has an infinite number of solutions:
5x + 2y = 2k,
2(k + 1)x + ky = (3k + 4).
Solution
The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y - 2k = 0 ….(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k + 1)x + ky - (3k + 4) = 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 = 5, b_1= 2, c_1= -2k and a_2 = 2(k + 1), b_2 = k, c_2= -(3k + 4)`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`5/(2(k+1)) = 2/k = (−2k)/(−(3k+4))`
⇒52(𝑘+1) = 2𝑘 = 2𝑘(3𝑘+4)
Now, we have the following three cases:
Case I:
`5/(2(k+1)) = 2/k`
⇒ 2 × 2(k + 1) = 5k
⇒ 4(k + 1) = 5k
⇒ 4k + 4 = 5k
⇒ k = 4
Case II:
`2/k = (2k(3k+4))`
⇒ `2k^2` = 2 × (3k + 4)
`⇒ 2k^2 = 6k + 8 ⇒ 2k^2 – 6k – 8 = 0`
`⇒ 2(k^2 – 3k – 4) = 0`
⇒ k2 – 4k + k – 4 = 0
⇒ k(k – 4) + 1(k – 4) = 0
⇒ (k + 1) (k – 4) = 0
⇒ (k + 1) = 0 or (k – 4) = 0
⇒ k = -1 or k = 4
Case III:
`5/(2(k+1)) = (2k)/((3k+4))`
`⇒ 15k + 20 = 4k^2 + 4k`
`⇒ 4k^2 – 11k – 20 = 0`
`⇒ 4k^2– 16k + 5k – 20 = 0`
⇒ 4k(k – 4) + 5(k – 4) = 0
⇒ (k – 4) (4k + 5) = 0
`⇒ k = 4 or k = (−5)/4`
Hence, the given system of equations has an infinite number of solutions when k is equal to 4.
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