KeerthanaPosted on The value of k, for which the system of equations 3x – ky – 20 = 0 and 6x – 10y + 40 = 0 has no solution, a3 b5 c6 d10 Answer : Option BExplanation : System of equations a1x+b1y+c1=0a_{1} x+b_{1} y+c_{1}=0a1x+b1y+c1=0 and a2x+b2y+c2=0a_{2} x+b_{2} y+c_{2}=0a2x+b2y+c2=0 will have no solution if a1a2=b1b2≠c1c2frac{a_{1}}{a_{2}}=frac{b_{1}}{b_{2}} eq frac{c_{1}}{c_{2}}a2a1=b2b1=c2c1 ∴36=−k−10≠−2040 herefore frac{3}{6}=frac{-k}{-10} eq frac{-20}{40}∴63=−10−k=40−20 ⇒12=k10⇒k=102=5Rightarrow frac{1}{2}=frac{k}{10} Rightarrow k=frac{10}{2}=5⇒21=10k⇒k=210=5 Rate This:NaN / 5 - 1 votesAdd comment