<PRE><B>Question 781429</B>
6c-1 = 0&lt;P&gt;
6c=1&lt;P&gt;
c=1/6&lt;P&gt;
For absolute value, you also have to consider the negative of what&#39;s inside the bars, because |x| = x if x is positive, and -x if x is negative.&lt;P&gt;
-(6c-1) = 0&lt;P&gt;
-6c +1 = 0&lt;P&gt;
-6c=-1&lt;P&gt;
c = 1/6&lt;P&gt;
When the absolute value is = 0, the solution will always be the same, because the graph of |ac+y| is a v that meets at y=0&lt;P&gt;
{{{ graph( 300, 200, -5, 5, -10, 10, abs(6x-1)) }}}</PRE>