Solving Compound Inequalities | MathHelp.com - By MathHelp.com
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| 00:00 | before graphing the combined inequality that you see here . | |
| 00:04 | We need to solve each of the two individual inequalities | |
| 00:09 | for X . So let's start with the one on | |
| 00:12 | the left . Our first step would be to distribute | |
| 00:15 | this negative three through the parentheses , which gives us | |
| 00:19 | 5 -6 . X -3 is greater than or equal | |
| 00:25 | to 14 . Simplifying the left side . One step | |
| 00:29 | further , 5 -3 is two , so we have | |
| 00:33 | to -6 . X is greater than or equal to | |
| 00:38 | 14 , isolating the X term . We subtract two | |
| 00:42 | from both sides and negative six X is greater than | |
| 00:48 | or equal to 12 . But watch out for your | |
| 00:52 | last step To get X by itself . We would | |
| 00:55 | divide both sides by -6 . But remember if you | |
| 01:01 | divide both sides of an inequality by a negative number | |
| 01:05 | , you must switch the direction of the inequality sign | |
| 01:09 | . So we have X is less than or equal | |
| 01:14 | to negative two over on the right . Are first | |
| 01:19 | step would be to combine the X terms on the | |
| 01:22 | left side of the inequality By subtracting four x from | |
| 01:28 | both sides . That leaves us with five , X | |
| 01:33 | is less than 15 And we divide both sides by | |
| 01:38 | five , So X is less than three . So | |
| 01:43 | we have X is less than or equal to negative | |
| 01:46 | two or X is less than three . Now we | |
| 01:55 | have a situation that looks just like the problems in | |
| 01:58 | the previous section . Our next step would be to | |
| 02:02 | graph each of these two inequalities just above the number | |
| 02:05 | line so that we can see what's going on before | |
| 02:08 | we try to combine them . Her ex is less | |
| 02:11 | than or equal to -2 . We have a closed | |
| 02:14 | dot At -2 and an arrow going to the left | |
| 02:19 | Rex is less than three . We have an open | |
| 02:22 | dot at three and an arrow going to the left | |
| 02:27 | . Remember to use long arrows , remember that or | |
| 02:32 | represents union and the union of these two inequalities means | |
| 02:39 | that our answer will include everything that's mentioned in the | |
| 02:43 | two inequalities . So all we have to do is | |
| 02:46 | bring both of these graphs down to the number line | |
| 02:50 | and see what we're left with . So we have | |
| 02:55 | X is less than or equal to negative two along | |
| 03:00 | with X is less than three . Mhm . If | |
| 03:10 | you take a look at what we're left with , | |
| 03:12 | we have the graph of X is less than three | |
| 03:16 | on the number line , so our answer reads , | |
| 03:20 | the set of all X is such that X is | |
| 03:25 | less than three . |
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