Skip to main contentSkip to solution

The value of k, for which the system of equations 3x-ky-20 = 0, and 6x-10y+40 = 0 has no solution, is:

Solution

✅ Correct Option: 3

For two linear equations to have no solution, the lines must be parallel.

For equations:

  • a1x+b1y+c1=0a_1x + b_1y + c_1 = 0
  • a2x+b2y+c2=0a_2x + b_2y + c_2 = 0

The condition for no solution is:

a1a2=b1b2≠c1c2\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}


From 3x−ky−20=03x - ky - 20 = 0:

a1=3a_1 = 3

b1=−kb_1 = -k

c1=−20c_1 = -20

From 6x−10y+40=06x - 10y + 40 = 0:

a2=6a_2 = 6

b2=−10b_2 = -10

c2=40c_2 = 40


Calculate a1a2\dfrac{a_1}{a_2}:

a1a2=36=12\dfrac{a_1}{a_2} = \dfrac{3}{6} = \dfrac{1}{2}


For no solution, b1b2=a1a2\dfrac{b_1}{b_2} = \dfrac{a_1}{a_2}:

b1b2=−k−10=k10\dfrac{b_1}{b_2} = \dfrac{-k}{-10} = \dfrac{k}{10}

Setting equal to a1a2\dfrac{a_1}{a_2}:

k10=12\dfrac{k}{10} = \dfrac{1}{2}

k=10×12k = 10 \times \dfrac{1}{2}

k=5k = 5


Check c1c2\dfrac{c_1}{c_2}:

c1c2=−2040=−12\dfrac{c_1}{c_2} = \dfrac{-20}{40} = -\dfrac{1}{2}

Since 12=12≠−12\dfrac{1}{2} = \dfrac{1}{2} \neq -\dfrac{1}{2}, the condition for no solution is satisfied.

Therefore, k=5k = 5

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question