For what value of $k$ does the line represented by the equation $-\frac{1}{2}-2kx = 5y$ contain the point $\left(\frac{1}{4},-6\right)$?
\(-\frac{1}{2}-2kx = 5y \)
\( \left(\frac{1}{4},-6\right)\)
So we can solve this for "k'
-1/2 - 2k(1/4) = 5(-6)
-1/2 - (1/2)k = - 30 multiply through by 2
-1 - k = -60
-k = - 59
k = 59
+612 
