Graphing Systems of Equations | MathHelp.com - By MathHelp.com
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| 00:0-1 | here were asked to solve this system of equations by | |
| 00:03 | graphing . So we're looking for the point of intersection | |
| 00:07 | for these two lines , there are a couple differences | |
| 00:11 | with this problem . However , first of all notice | |
| 00:14 | that our two lines are not written in Y equals | |
| 00:18 | mx plus B form . So our first task is | |
| 00:22 | to convert these equations to the proper form . Before | |
| 00:26 | graphing . For our first equation , we would subtract | |
| 00:32 | three X from both sides to get negative Y equals | |
| 00:36 | negative three , X minus three , Divide both sides | |
| 00:41 | by -1 and Y equals three . X plus three | |
| 00:49 | . For our second equation subtract X from both sides | |
| 00:54 | to get negative two , Y equals negative X plus | |
| 00:59 | four , Divide both sides by -2 and Y equals | |
| 01:07 | positive one half x minus two . Now we're ready | |
| 01:14 | to graph our lines For our first line , we | |
| 01:18 | have a y intercept of three And think of our | |
| 01:23 | slope as 3/1 . So we go up three and | |
| 01:28 | to the right one to plot a second point before | |
| 01:34 | graphing this line . However , let's plot a couple | |
| 01:37 | points on our second line so that we can determine | |
| 01:41 | the general neighbourhood of where our two lines are gonna | |
| 01:45 | intersect before we actually draw them . For our second | |
| 01:50 | equation , the y intercept is negative two and our | |
| 01:55 | slope is one half , so we go up one | |
| 01:58 | and to the right to To plot a 2nd point | |
| 02:03 | . Notice that our two lines are gonna intersect somewhere | |
| 02:07 | to the left of the y axis , so we | |
| 02:12 | need points to the left of the y axis instead | |
| 02:15 | of the right . So for our first equation , | |
| 02:19 | for example , instead of using our slope of 3/1 | |
| 02:24 | to go up three and to the right one to | |
| 02:26 | plot a point . Let's go in the opposite direction | |
| 02:29 | with our slope Down three and to the left one | |
| 02:34 | to plot a point . If we go down three | |
| 02:38 | into the left one again , you see that we | |
| 02:42 | have a point that might be a candidate for a | |
| 02:46 | point of intersection of the two lines for our second | |
| 02:50 | line are slope of one half can move us up | |
| 02:54 | one into the right to , but it can also | |
| 02:56 | move us down one and to the left to . | |
| 03:00 | So this is actually our point of intersection . We | |
| 03:05 | graph are two lines . Mhm . Oh you can | |
| 03:13 | see that our point of intersection has the coordinates negative | |
| 03:17 | to negative three . So that's the solution to our | |
| 03:23 | system . |
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