Systems of Inequalities | MathHelp.com - By MathHelp.com
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| 00:0-1 | to solve the given system of inequalities . We start | |
| 00:03 | by graphing the associated equation for each inequality . In | |
| 00:09 | other words , we graph y equals negative 1/5 X | |
| 00:13 | plus one , and Y equals three X plus two | |
| 00:19 | . So for the first inequality we start with our | |
| 00:23 | y intercept of positive one Up one Unit on the | |
| 00:27 | Y Axis . From there we take our slope of | |
| 00:31 | negative 1/5 So we go down one and to the | |
| 00:35 | right five and plot a second point . Now notice | |
| 00:40 | that our inequality uses a less than sign . This | |
| 00:45 | means that we draw a dotted line connecting the points | |
| 00:49 | rather than a solid line . It's important to understand | |
| 00:54 | that if we have a greater than sign or a | |
| 00:56 | less than sign , we use a dotted line and | |
| 01:01 | if we have a greater than or equal to sign | |
| 01:03 | or less than or equal to sign , we use | |
| 01:07 | a solid line . Pay close attention to this idea | |
| 01:12 | when drawing your lines , Students often carelessly use a | |
| 01:17 | solid line when they should use a dotted one and | |
| 01:21 | vice versa . Next let's take a look at our | |
| 01:26 | second inequality , which has a Y intercept of positive | |
| 01:30 | . Too Up two units on the y axis . | |
| 01:34 | From there we take our slope of three or 3/1 | |
| 01:39 | , so we go up three and to the right | |
| 01:42 | one and plot a second point and notice that this | |
| 01:46 | inequality uses a greater than or equal to sign . | |
| 01:52 | So we connect the points with a solid line rather | |
| 01:56 | than a dotted line . Next , we need to | |
| 01:59 | determine which side of each of these lines to shade | |
| 02:04 | on the graph to determine which side of our first | |
| 02:09 | line . To shade . We use a test point | |
| 02:12 | on either side of the first line . The easiest | |
| 02:16 | test point to use is 00 . So we plug | |
| 02:20 | zero into our first inequality for both X and Y | |
| 02:25 | And we have zero is less than negative . 1 | |
| 02:29 | 5th Time zero Plus 1 , which simplifies to zero | |
| 02:35 | is less than zero plus one Or zero is less | |
| 02:39 | than one . Notice that zero is less than one | |
| 02:44 | is a true statement . This means that our test | |
| 02:48 | zero is a solution to the first inequality . So | |
| 02:54 | we shade in the direction of 00 along our first | |
| 02:59 | boundary line . Next we determine which side of our | |
| 03:04 | second line . To shade By using a test point | |
| 03:07 | on either side of the 2nd line , Such as | |
| 03:11 | 00 . Plugging a zero into our second inequality for | |
| 03:17 | both X and Y , we have zero is greater | |
| 03:21 | than or equal to three times zero plus two , | |
| 03:26 | which simplifies to zero is greater than or equal to | |
| 03:29 | zero plus two , Or zero is greater than or | |
| 03:33 | equal to two . Notice that zero is greater than | |
| 03:37 | or equal to two is a false statement . This | |
| 03:41 | means that our test zero is not a solution to | |
| 03:46 | the inequality , So we shade away from 00 along | |
| 03:52 | our second boundary line . Finally , it's important to | |
| 03:57 | understand that the solution to this system of inequalities is | |
| 04:01 | represented by the part of the graph where the two | |
| 04:05 | shaded regions overlap , which in this case is in | |
| 04:09 | the lower left . In other words , any point | |
| 04:14 | that lies in this part of the graph is a | |
| 04:16 | solution to the given system of inequalities . Note that | |
| 04:21 | the points along the dotted boundary line of this region | |
| 04:25 | are not solutions to the system , but the points | |
| 04:29 | along the solid boundary line of this region are solutions | |
| 04:33 | to the system . |
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