Addition elimination method 3 | Systems of equations | 8th grade | Khan Academy - By Khan Academy
Transcript
| 00:0-1 | solve for X and Y , and we have the | |
| 00:01 | system of equations right here we have two x minus | |
| 00:04 | y is equal to 14 and negative six x plus | |
| 00:07 | three y is equal to negative 42 so we could | |
| 00:10 | try to solve this by elimination . And maybe we | |
| 00:13 | want to . Let's see if we can eliminate . | |
| 00:16 | Let's see if we can eliminate ry variables first . | |
| 00:18 | We have a three y here and we have a | |
| 00:20 | negative y up here and we can . They won't | |
| 00:23 | eliminate . If you just had negative y plus three | |
| 00:26 | y , that won't eliminate . But if we could | |
| 00:28 | turn this negative y into a negative three y that | |
| 00:30 | it would cancel out with a three y and the | |
| 00:32 | best way to turn a negative Y into a negative | |
| 00:34 | three y is to multiply this entire top equation by | |
| 00:38 | three . So let's do that . Let's multiply . | |
| 00:40 | Let me get some space over to the left . | |
| 00:42 | Let's multiply this entire top equation by three so I'm | |
| 00:46 | gonna multiply it by three . So I multiply two | |
| 00:49 | x by three . I get six x , I | |
| 00:51 | multiply negative y by three . I get negative three | |
| 00:54 | y and then I multiply 14 by 3 33 times | |
| 01:00 | 14 is 42 right , three times 10 is 30 | |
| 01:05 | plus 12 . It's 42 and then we can add | |
| 01:08 | both of these equations . Something interesting should already maybe | |
| 01:11 | be showing up on your radar . Let's add both | |
| 01:14 | of these equations . Let's add the left hand side | |
| 01:17 | . Negative six X plus six x Well , those | |
| 01:19 | cancel out . We get zero . Then we have | |
| 01:21 | three y minus three wide . Those cancel out , | |
| 01:24 | we get another zero . And then finally you get | |
| 01:27 | negative . 42 plus 42 . Well , that's zero | |
| 01:30 | . So we end up with just zero is equal | |
| 01:33 | to zero , which is clearly true . But it's | |
| 01:35 | not putting any constraints on the X or Y . | |
| 01:38 | And that's because whenever you have a situation like this | |
| 01:40 | where you just get something that's obviously true , zero | |
| 01:43 | equals zero or one equals one or five equals five | |
| 01:46 | . What we're dealing with a situation is where both | |
| 01:49 | of our constraints , both of our equations , are | |
| 01:51 | actually the same equation . So this right here is | |
| 01:54 | a dependent system . It is a dependent . It | |
| 01:59 | is a dependent system and you see it right over | |
| 02:01 | here . If you take that first equation you multiply | |
| 02:04 | it by three . You got six X minus three | |
| 02:06 | Y is equal to 42 if we then multiply it | |
| 02:09 | by negative one . If we then multiplied it by | |
| 02:11 | negative one , you would get this . You would | |
| 02:14 | get the exact same equation as the second equation . | |
| 02:17 | You would get negative six X plus three y is | |
| 02:21 | equal to negative 42 or another way to think about | |
| 02:24 | it . If you want to go from the first | |
| 02:25 | equation to the second equation , you just multiply both | |
| 02:28 | sides of the equation . Times , times negative three | |
| 02:33 | . So both of these constraints are actually the same | |
| 02:35 | constraints . They're just kind of a scaled up multiple | |
| 02:37 | of each other . So if you were to graph | |
| 02:39 | them , and I might as well grab them for | |
| 02:41 | you right here , this first equation right here let | |
| 02:44 | me do it . Over here , it's two x | |
| 02:46 | minus . Y is equal to 14 . Could subtract | |
| 02:49 | two x from both sides , and you would get | |
| 02:52 | Let me just subtracted . Subtract two x , subtract | |
| 02:55 | two X on the left hand side . You're left | |
| 02:57 | with just negative y On the right hand side , | |
| 03:00 | you have negative two X plus 14 . Multiply both | |
| 03:03 | sides by negative one , and you get Y is | |
| 03:06 | equal to two X minus 14 . So this first | |
| 03:10 | equation over here , if I would draw my axes | |
| 03:14 | . So that is my Y axis . And then | |
| 03:18 | that is my X axis . This graph right over | |
| 03:22 | here , it's why intercept is negative . 14 . | |
| 03:25 | So this is zero negative 14 0 negative 14 and | |
| 03:29 | has a slope of two . So it's going to | |
| 03:31 | look something like this . It's going to look something | |
| 03:33 | like that . And then the second equation , if | |
| 03:36 | you were to graph it because this is a dependent | |
| 03:38 | system is the exact same line . If you were | |
| 03:41 | to put this in slope intercept form and graph it | |
| 03:43 | , you would get the exact same thing you would | |
| 03:45 | would go right on top of it . So there's | |
| 03:47 | actually an infinite number of solutions . These two lines | |
| 03:50 | are the same lines , so they intersect everywhere on | |
| 03:54 | each of the respective lines . They're the same line | |
| 03:56 | , and that's where and when you get something like | |
| 03:58 | this , zero equals zero or one equals one . | |
| 04:01 | That's the tell tale sign that you're dealing with a | |
| 04:04 | dependent system . If you've got something like zero equals | |
| 04:07 | one , then you would have no solution , so | |
| 04:10 | this would right here would be an inconsistent in consistence | |
| 04:15 | . This would be an inconsistent system , and this | |
| 04:17 | would be the situation . So let me make it | |
| 04:19 | very clear this is a situation we're dealing with the | |
| 04:21 | same same lines , the lines of the same lines | |
| 04:26 | . This is a situation where you have parallel lines | |
| 04:29 | , parallel parallel lines and so they never intersect . | |
| 04:34 | And then obviously you have the the easiest situation . | |
| 04:37 | I think this is one most of us familiar with | |
| 04:39 | where you have something like X is equal to one | |
| 04:41 | , or why is equal to two or anything like | |
| 04:43 | this doesn't have to be one or two . And | |
| 04:45 | then this is clearly a situation where you have an | |
| 04:48 | independent system where you have two different lines that intersect | |
| 04:51 | in exactly one place anyway . This there's an infinite | |
| 04:55 | number of solutions . Any X and Y that satisfy | |
| 04:58 | the first equation will also satisfy the second equation because | |
| 05:01 | there is fundamentally the same constraint |
Summarizer
DESCRIPTION:
Khan Academy presents Addition Elimination Method 3, an educational video resource on math.
OVERVIEW:
Addition elimination method 3 | Systems of equations | 8th grade | Khan Academy is a free educational video by Khan Academy.It helps students in grades 9,10,11,12 practice the following standards HSA.REI.C.5,HSA.REI.C.6,.
This page not only allows students and teachers view Addition elimination method 3 | Systems of equations | 8th grade | Khan Academy but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.
1. HSA.REI.C.5 : Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
2. HSA.REI.C.6 : Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables.
GRADES:
9
10
11
12
STANDARDS:
HSA.REI.C.5
HSA.REI.C.6
