Question

For what value of k, $x-3y=7$ and $kx+6y=5$ will have no solution?

Hint:

For a system of linear equations to have no solution, the lines must be parallel, which occurs when the ratios of the coefficients of $x$ and $y$ are equal.

Answer 1

Step 1: Recognize the condition for no solution.
For the system of linear equations to have no solution, the lines must be parallel. The condition for two lines to be parallel is that the ratio of the coefficients of $x$ and $y$ in both equations must be equal.
Step 2: Apply the condition.
The first equation is $x - 3y = 7$, and the second equation is $kx + 6y = 5$. The condition for parallel lines is: \[ \frac{1}{k} = \frac{-3}{6} \]
Step 3: Solve for $k$.
Solving the above equation gives: \[ \frac{1}{k} = \frac{-1}{2} \quad \Rightarrow \quad k = -2 \]
Step 4: Conclusion.
Thus, the value of $k$ for which the system has no solution is $k = -2$.