Solving Systems of Equations by Substitution | MathHelp.com - By MathHelp.com
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00:0-1 | to solve this system of equations by substitution . The | |
00:05 | idea is simple since our first equation tells us that | |
00:11 | Y equals three X plus two , we can substitute | |
00:17 | a three X plus two in for the Y . | |
00:20 | In our second equation . In other words , since | |
00:24 | why means the same thing as three X plus two | |
00:32 | . We can replace the Y . In the second | |
00:34 | equation with a three X plus two . Our second | |
00:41 | equation then becomes seven X minus four times parentheses . | |
00:49 | Three X plus two equals seven . Now we can | |
00:55 | solve for X . If we simplify on the left | |
00:59 | side , we get seven X minus 12 , X | |
01:04 | minus eight equals seven , Which simplifies to negative five | |
01:09 | , X -8 equals seven At 8 to both sides | |
01:16 | , and negative five X equals 15 , divide both | |
01:21 | sides by negative five and X equals negative three . | |
01:28 | To find why Plug a negative three back in for | |
01:33 | X . In the first equation to get y equals | |
01:40 | three times negative three plus two , which simplifies to | |
01:47 | Y equals negative nine plus two , Or y equals | |
01:52 | -7 . So the solution to this system of equations | |
01:57 | is -3 -7 . |
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