Multiple choice

For what value(s) of k does the given system of equations have no solution?

kx + 2y = 3 3x + 6y = 10

  1. k = 3/10

  2. k = 1

  3. Either k = 3/10 or k = 1

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

kx + 2y = 3 3x + 6y = 10 Comparing above equations with  a1x + b1y = c1 a2x + b2y = c2 We get $\frac{a_1}{a_2} = \frac{k}{3}$ , $\frac{b_1}{b_2} = \frac{2}{6}$ , $\frac{c_1}{c_2} = \frac{3}{10}$ For no solution: a1/a2  = b1/b2 $ \neq$ c1/c2 $\Rightarrow$ k/3  = 2/6 $ \neq$ 3/10 $\Rightarrow$ k = 1 $ \neq$ 9/10 The given system of equations will have no solution if k  =  1.  

AI explanation

A pair of linear equations a1x+b1y=c1 and a2x+b2y=c2 has no solution when a1/a2 = b1/b2 ≠ c1/c2 (parallel, non-coincident lines). Here k/3 = 2/6 = 1/3, so k = 1. Checking the constant ratio, 3/10 is not equal to 1/3, confirming the lines are parallel with no common solution when k=1.

Keep practicing — related questions