+44 (3−k)x−4 (3−k)y=−27
In the given linear equation, kk is a constant and k≠3. When graphed in the xy-plane, what is the slope of the line?
I think the slope is 3/12 but I'm not sure.
the slope would be 1/4
To determine the slope of the graph of this linear equation, the equation can be manipulated into the slope-intercept form, y=mx+by=mx+b, by solving for yy. First, subtract (3−k)x(3−k)x from both sides of the equation to get −4(3−k)y=−(3−k)x−27−4(3−k)y=−(3−k)x−27. Then divide both sides by −4(3−k)−4(3−k) to get y=−(3−k)x−27−4(3−k)=−(3−k)x−4(3−k)−27−4(3−k)=14x+274(3−k)y=−(3−k)x−27−4(3−k)=−(3−k)x−4(3−k)−27−4(3−k)=14x+274(3−k). The slope is the coefficient of the xx term, or 1414.
