MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Graphing a system of inequalities requires graphing them both on the same coordinate plane to see where the values overlap in order to determine the solutions. When graphing inequalities, remember to use the same rules as you would when graphing a line. If the inequalities are already solved for y, start by plotting the y intercept (the constant). From there, use the slope (the coefficient of the x variable) to plot a second point. If the inequality is less than or equal to, or greater than or equal to, then draw a solid line. If the inequality is less than or greater than (but not equal to), draw a dotted line. To determine the region that should be shaded, choose a point anywhere on the coordinate plane and substitute the values in the inequality. If the inequality is true, then shade the region where the point was taken from. If the inequality if false, then shade the region on the other side of the inequality line. Take the same steps with the other inequality of the system. The area where the shaded regions of the two inequalities overlap is the solution.
Here I'm asked to graph a system of inequalities. What that means is I have two inequalities that both have two variables. I'm going to graph them one by one and then look for the solution region which is going to be where my shadings overlap.
This first one I'm going to graph in blue, by the way colour pencil help with these problems because you are going to be doing two different shadings.
Okay so the first thing I'm going to do is graph that line using my slope intercept techniques because its already in y equals mx plus b form. My first dot will go at the y intercept from there I'm going to count the slope which is up three over one. If you have graph paper your will be really precise.
The next thing I'm going to do is graph this line as a dashed line because this is just a greater than. If it were greater than an equal to I would use a solid line. Okay so there's my dashed line. This by the way is why you want to make sure you're using pencil. So you don't have to erase your dashes with pen.
Okay next thing I need to do still on this first line is figure out which way to shade. The way to figure that out is to pick a point substitute in your x and y coordinate pair and see if you get a true or false inequality. I usually pick (0,0) as long as its not on the line.
So let's see is it true that 0 is greater than 3 time 0 take away 2? Yeah right, 0 is greater than -2. That means I do want to shade the region that includes that (0,0) point. Here we go, shading, okay there we go. There is my blue line with all the shading and dashing is done.
Now I'm going to do another one in red I'm going to have to do my y intercept of +4, see how these graphs get really ugly really quick. From there I need to do the slope which is down one, over two, something like that.
This line is going to be a solid line because it has that equal to quantity or equal to property. Okay sound effects are optional okay next thing I'm going to do is pick a point and substitute it in. I again like to use (0,0). Is it true that 0 is less than or equal to half of 0 plus 4? Let's see 0 is less than or equal to 4 yeah that's true. That means from my red line I'm going to shade this region.
Okay to answer the problem I need to find the solution region its going to be the piece of my graph for all of my shadings overlap. And this is why it gets really tricky if you are not suing coloured pencils.
The place where my two shadings overlap is all the way up to that red solid line and then stopping at that blue dashy line. It's this region here, what a mess. I feel like a little kindergartener colouring with my crayons.
This is a correct mathematical problem my solution region is where my two shadings overlap. And you can see it because I tried to darken it using black. I just wanted to make sure I was really careful to stop at this boundary line and stop at that boundary line when I was shading.
So when you come to a system of inequalities graph them each individually being careful with solid lines and dashy lines make sure you do the shadings and then darken where your two shaded regions overlap.
Unit
Solving Systems of Equations