Find the value of k so that the system of equation has no solution 3x - y - 5 = 0 6x - 2y + k = 0

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Solution

Short Answer

Here, a1 = 3, b1 = -1 and c1 = -5
a2 = 6, b2 = -2 and c2 = k

The given system of equations:3x - y - 5 = 0                      ….(i)And, 6x - 2y + k = 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 = 3, b_1= -1, c_1= -5 and a_2 = 6, b_2= -2, c_2 = k`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`` i.e., 3/6 = (−1)/(−2) ≠ −5/k``⇒(−1)/(−2) ≠ (−5)/k ⇒ k ≠ -10`

Hence, equations (i) and (ii) will have no solution if k ≠ -10.

Page 2

The given system of equations can be written askx + 3y + 3 - k = 0                       ….(i)12x + ky - k = 0                           ….(ii)This system of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`

where, `a_1 = k, b_1= 3, c_1 = 3 - k and a_2 = 12, b_2 = k, c_2= –k`

For the given system of linear equations to have no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ k/12 = 3/k ≠ (3−k)/(−k)``⇒k/12 = 3/k and 3/k ≠ (3−k)/(−k)``⇒ k^2 = 36 and -3 ≠ 3 - k`⇒ k = ±6 and k ≠ 6⇒k = -6

Hence, k = -6.

Page 3

The given system of equations:5x - 3y = 0                 ….(i)2x + ky = 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= -3, c_1 = 0 and a_2 = 2, b_2 = k, c_2 = 0`For a non-zero solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2)``⇒ 5/2 = (−3)/k``⇒5k = -6 ⇒ k = (−6)/5`

Hence, the required value of k is `(−6)/5`.