Here it is again! I know how to do his,I just don't have the resourse, can someone help me to graph this problem. x - y < 3 x + 2y > 6 This is the graph of a shaded region First draw the graphs of the boundary lines. Boundary line's equations: | Boundary line's intercepts | x - y = 3 | (0,-3) (3,0) x + 2y = 6 | (0,3) (6,0) | --------------------------------------------------------- Just draw one boundary line at a time so you can tell which area to shade. Draw the first bondary line by plotting its two intercepts and drawing a dotted line thru them. Draw it dotted since it is strictly less than. I can only draw it solid on here, but you can draw it dotted on your paper:Now before drawing the other boundary line, we must decide which side of this line we will shad on. To do this we pick any point which is not on that line as a test point, say, we pick (3,2) for we can be sure that is not on the line. We substitute x=3 and y=2 into the inequality x - y < 3 3 - 2 < 3 1 < 3 That is a true inequality, since 1 really is less than 3, so we know to shade the side of the boundary line that the test point (3,2) is on. That's the side above the line. [BTW if that inequality had come out false, then we'd have known to shade the other side, the lower side. instead.] Now draw the second boundary line by plotting its two intercepts and drawing a dotted line thru them. Draw it dotted also since it is strictly greater than. I can only draw it solid on here, but you can draw it dotted on your paper: Now we must also decide which side of this line we will shade. As before, to do this we pick any point which is not on that line as a test point, say, we pick (2,1) for we can be sure that is not on the green line. We substitute x=2 and y=1 into the inequality x + 2y > 6 2 + 2(1) > 5 4 > 5 That is a false inequality, since 4 is NOT greater than 5, so we know to shade the side of the boundary line that the test point (2,1) is NOT on. That's the side above the green line. So the section we shade is the V-shaped region which is above both the red and the green lines. And the lines are supposed to be drawn dotten, not solid. [You would have drawn them solid if the inequalities had been < or >.] ============================================================== another problem that I have is, solve the system by addition or substitution. 3x + 6y = 0 x = 2/3 Since the second equation already gives you the value for x, namely 2/3, the problem is half-worked for you from the start. So all you have to do is finish by substituting 2/3 for x in the first equation and solve for y. 3x + 6y = 0 3(2/3) + 6y = 0 2 + 6y = 0 6y = -2 y = -2/6 y = -1/3 So the solution is (x,y) = (2/3,-1/3) Edwin