While working through the third of three packets I'm going through to review for a pre-test for an independent-study calculus class, I came across the following problem:
For what value of $k$ does the system of equations $\begin{cases}2x + y = 1 \\ 12x + ky = 12\end{cases}$ have no solution?
Could somebody give me an example of how to solve a problem like this without giving away the answer to this problem?
On
Say we have $ax+by=c ; mx+ny=p $ So that means if we solve each of them for y we have $y=\frac{-a}{b}x+c \text{ and } y=\frac{-m}{n}x+p $ If these lines are to be parallel (meaning they will never intersect meaning their system will not have a solution ), then we need $\frac{a}{b}=\frac{m}{n}$ and $c \neq p$.
Solve the equations for $x$ and $y$ in terms of $k$. Then you will see $k$ in a denominator somewhere. Pick $k$ so that the denominator is zero.