We need to graph this equation:

Its solutions are the points through which it graph passes. Since it's a linear equation its graph is a straight line and we only need two of its points to draw it. But before graphing let's re-write the equation. We can substract 16x from both sides:

And we divide both sides by 2:

So now with this equation if we pick two random x values we'll get their corresponding y values. This way we'll find two points that are part of the graph which is the line that passes through both. We can begin with x=0:

So the first point is (0,150). Then we can take x=10 and we get:

So the second point is (10,70). Then the graph is the line that passes through points (0,150) and (10,70). In order to represent it