Math Tutoring | MathHelp.com - By MathHelp.com
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| 00:0-1 | in this problem are variable X appears on both sides | |
| 00:05 | of the equation . In this situation , in order | |
| 00:09 | to solve for X . Our goal is to get | |
| 00:12 | our X terms together on one side of the equation | |
| 00:16 | and leave our number on the other side of the | |
| 00:18 | equation . So how do we put our terms 10 | |
| 00:22 | X . And three X together If they appear on | |
| 00:26 | different sides of the equation ? Well , notice that | |
| 00:30 | if we subtract three X from both sides of the | |
| 00:34 | equation . Yeah , on the left side , 10 | |
| 00:39 | X minus three X . Simplifies to seven X . | |
| 00:44 | Yeah . And on the right side are three X | |
| 00:47 | -3 x cancels out . And we're left with 56 | |
| 00:54 | . Now we have the situation we're looking for with | |
| 00:57 | our X terms together on one side of the equation | |
| 01:01 | and our number on the other . And all we | |
| 01:04 | have to do to solve for X is divide both | |
| 01:07 | sides of the equation by seven And we have x | |
| 01:14 | equals eight . Mhm . Note that we can check | |
| 01:20 | our answer by plugging eight back in for both of | |
| 01:24 | the X is in the original equation . So go | |
| 01:28 | ahead and check the answer and see if it works | |
| 00:0-1 | . |
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