Saved Bookmarks
| 1. |
The system x + 2y = 3 and 5x + ky + 7 = 0 has no solution for what value of k? |
|
Answer» Given system of equation is x + 2y = 3 ⇒ x + 2y – 3 = 0 And 5x + ky + 7 = 0. By comparing given system of equations with a1x + b1y + c1 = 0 & a2x + b2y + c2 = 0, We get a1 = 1, b1 = 2, c1 = –3 and a2 = 5, b2 = k, c2 = 7. Since, given that system of equations has no solution. \(\therefore \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}.\) Therefore, \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\) & \(\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) ∴ \(\frac{1}{5}\) = \(\frac{2}{k}\)⇒ k = 2× 5 = 10. (By cross multiplication) And \(\frac{2}{k}\) ≠ \(\frac{-3}{7}\) ⇒ –3k ≠14 ⇒ k ≠ \(\frac{-14}{3}\) . Hence, for k = 10, the given system of equations has no solution. |
|