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Systems of Inequalities  Problem 2
Alissa Fong
Alissa Fong
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
When graphing a system of linear inequalities, first make sure that the inequalities are in "graphable" form, meaning that the y variable is isolated and everything else is on the other side of the equal sign (the inequality equivalent to slopeintercept y=mx+b form). If it is not, use inverse operations to isolate the y variable. To plot points, use the same rules you would when graphing a line  plot the yintercept and from there, use the slope 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 have a system of inequalities where one of my equations is not in y equals mx plus b form and the other one is going to be a horizontal line. So it will a little bit different from what you might have seen before. I'm going to graph each one and then look for where the shading overlaps.
First I'm going to do my red line but I need to solve it for y. I have 2y is greater than 4x plus 6. In order to get y isolated I need to divide by 2. Ohoh! Dividing by a negative, I try to be dramatic because that means I'm going to have to change the direction of my inequality sign. Instead of being greater than now it's going to look like that, I'm going to have a 2x minus 3. That's the line I'm going to graph. Starting with the y intercept I'll go up 2 over 1 to represent the slope graph that guy with a dashy line. Now I'm ready to think about shading.
In order to do my shading I need to pick up a point that's not on the line and substitute it in probably to my original just in case I made any mistakes. I usually use (0,0). Is it true that 0 is greater than4 times 0 plus 6? Is it true that 0 is greater than 6?No that means don't shade this half of the line because (0,0) was a no. Instead I'm going to shade over here.
Red line done, now I'm going to do the blue line I'm only half way done. The blue line is the horizontal line y equals 5 in a dashy form, 1, 2, 3, 4, 5, dashy okay. Now I want to shade y values that are greater than that. So that's going to be everything above 5.
Finally I'm ready to darken my solution region. The solution region is where my two shadings overlap so it's this area up here bounded by my diagonal red line and my horizontal blue line. This one was a little bit tricky because this equation was not in y equals mx plus b form and because I had a horizontal line but you just use the same steps as you would for any other systems of inequalities.
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Alissa Fong
M.A. in Secondary Mathematics, Stanford University
B.S., Stanford University
Alissa has a quirky sense of humor and a relatable personality that make it easy for students to pay attention and understand the material. She has all the math tips and tricks students are looking for.
Your tutorials are good and you have a personality as well. I hope you have more advanced college level stuff, because I like the way you teach.”
Thanks alot for such great lectures... I never found learning this easier ever before... keep up the great work.... :)”
You seem so kind, it's awesome. Easier to learn from people who seem to be rooting for ya!' thanks”
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Sample Problems (9)
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Problem 2 5,181 viewsGraph:
2y > 4x + 6y > 5 
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Problem 3 4,124 viewsGraph:
y > ¾x − 2y < ¾x − 3 
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a) Write a system of inequalitiesb) Graph the system to show all possible solutions. 
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Problem 5 127 views 
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Problem 6 101 views 
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Problem 7 112 views 
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Problem 8 117 views 
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