Question

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

Answer 1

The given system of equations:
2x + 3y = 7,
⇒ 2x + 3y - 7 = 0 ….(i)
And, (k – 1)x + (k + 2)y = 3k
⇒(k – 1)x + (k + 2)y - 3k = 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= 3, c_1= -7 and a_2 = (k – 1), b_2 = (k + 2), c_2= -3k`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`2/((c−1)) = 3/((k+2)) = (−7)/(−3k)`
`⇒2/((k−1)) = 3/((k+2)) = 7/(3k)`
Now, we have the following three cases:
Case I:
`2/((k−1)) = 3/(k+2)`
⇒ 2(k + 2) = 3(k – 1) ⇒ 2k + 4 = 3k – 3 ⇒ k = 7
Case II:
`3/((k+2)) = 7/(3k)`
⇒ 7(k + 2) = 9k ⇒ 7k + 14 = 9k ⇒ 2k = 14 ⇒ k = 7
Case III:
`2/((k−1)) = 7/(3k)`
⇒ 7k – 7 = 6k ⇒ k = 7
Hence, the given system of equations has an infinite number of solutions when k is equal to 7.