Addition elimination method 3 | Systems of equations | 8th grade | Khan Academy - Free Educational videos for Students in K-12 | Lumos Learning

Addition elimination method 3 | Systems of equations | 8th grade | Khan Academy - Free Educational videos for Students in k-12


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

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