Question

Problem

\[\begin{aligned}
-&x + \dfrac 1 2 y = -3\\\\
&0 = cx - 2y + 10
\end{aligned}\]
In the system of equations, \[c\] is a constant. For what value of \[c\] does the system of linear equations have no solutions?

Answer 1

Rewrite both equations in slope-intercept form.

First: -x + (1/2)y = -3 → multiply by 2: -2x + y = -6 → y = 2x - 6.

Second: 0 = cx - 2y + 10 → -2y = -cx - 10 → y = (c/2)x + 5.

For no solutions the lines must be parallel but distinct, so their slopes must be equal: 2 = c/2 → c = 4.