Question 68

Let k be a constant. The equations $$kx + y = 3$$ and $$4x + ky = 4$$ have a unique solution if and only if


Two linear equations ax+by= c and dx+ ey = f have a unique solution if $$\frac{a}{d}\ne\ \frac{b}{e}$$

Therefore, $$\frac{k}{4}\ne\ \frac{1}{k}$$ => $$k^2\ne\ 4$$

=> k $$\ne\ $$ |2|

Video Solution


Create a FREE account and get:

  • All Quant CAT complete Formulas and shortcuts PDF
  • 35+ CAT previous papers with video solutions PDF
  • 5000+ Topic-wise Previous year CAT Solved Questions for Free


Boost your Prep!

Download App