Solving Systems of Equations by Addition | MathHelp.com - By MathHelp.com
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| 00:0-1 | to solve this system of equations by addition , our | |
| 00:06 | first goal is to cancel out one of the variables | |
| 00:10 | by adding the two equations together notice that if we | |
| 00:16 | add these two equations together , R plus Y and | |
| 00:21 | minus Y will cancel out on the left three X | |
| 00:27 | plus seven X is 10 X . and on the | |
| 00:33 | right five plus 15 is 20 , Dividing both sides | |
| 00:39 | by 10 X equals two . To find why simply | |
| 00:49 | plug a to back in for X in either one | |
| 00:53 | of the two given equations . So let's go with | |
| 00:57 | our first equation . Three X plus Y equals five | |
| 01:02 | . If we plug it to back in for X | |
| 01:08 | , that gives us three times to Plus y equals | |
| 01:16 | five . This simplifies to six plus Y equals five | |
| 01:23 | . And subtracting six from both sides , gives us | |
| 01:28 | Y equals negative one , So x equals to n | |
| 01:35 | . y equals negative one . And put your answer | |
| 01:39 | in the form of the ordered pair , two negative | |
| 01:45 | one mm . |
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