Simultaneous Equations - the Elimination Method - How to solve - Math Lesson - By tecmath
Transcript
| 00:0-1 | Good day . Welcome to Tech Mouth channel . I'm | |
| 00:01 | josh in this video . We're going to have a | |
| 00:03 | look at the elimination method for solving simultaneous equations . | |
| 00:06 | So sit back and enjoy . I'm going to start | |
| 00:08 | out by putting some simultaneous equations up . Okay , | |
| 00:11 | so we'll start out with two X plus four Y | |
| 00:15 | . And this is going to equal 10 . That's | |
| 00:19 | our first equation . That we're going to have another | |
| 00:21 | equation where X plus nine , Y is equal to | |
| 00:27 | 12 . So as you'll notice a couple of things | |
| 00:30 | with simultaneous equations or get through the basics first . | |
| 00:33 | So first off , we have two sets of linear | |
| 00:35 | equations are these equations are made up of parts with | |
| 00:39 | letters in them . Okay , They letters are called | |
| 00:41 | variables . So we have the variable of X in | |
| 00:43 | both equations and we have the variables of why In | |
| 00:46 | both equations . In front of the variables we have | |
| 00:49 | these numbers which are called coefficients . So the coefficient | |
| 00:52 | here is we have a coefficient of two in front | |
| 00:53 | of the variable X . We have a coefficient of | |
| 00:56 | four in front of the variable of why in front | |
| 00:59 | of this one here , where it's not written , | |
| 01:01 | we assume that the actual coefficient here is one and | |
| 01:04 | it is going to be one . We just don't | |
| 01:06 | write it in . So that's probably what you want | |
| 01:08 | to get used to . First off , the basics | |
| 01:11 | are covered on this . So the way that you | |
| 01:13 | solve this particular set of equations , First off , | |
| 01:16 | I would go through and give these equations each name | |
| 01:19 | . This one here is going to be equation one | |
| 01:21 | and this is equation to this helps a bit later | |
| 01:24 | on when you're trying to identify what's happening to each | |
| 01:27 | equation as we go along . So the trick to | |
| 01:30 | doing these equations by the elimination method is you have | |
| 01:33 | to adjust the entire equations that we have here so | |
| 01:37 | that what we have is the coefficients in front of | |
| 01:39 | one of the variables here will match , I say | |
| 01:41 | . Either we end up with a two X here | |
| 01:43 | , or we end up with matching coefficients in front | |
| 01:45 | of the y variables here . Sounds complicated . It's | |
| 01:48 | not too bad . So I'm going to start off | |
| 01:50 | by having a look at the variables exhale , you're | |
| 01:53 | going to notice that we have to X and we | |
| 01:55 | have one exhale . So we can stuff around this | |
| 01:58 | entire equation , we can multiply its entire equation to | |
| 02:01 | here by two . And by doing that , we're | |
| 02:04 | going to end up with a two X . And | |
| 02:05 | then we'll have this matching uh coefficient in front of | |
| 02:09 | each of the variable X here . So let's do | |
| 02:11 | that . I'm gonna first off , I'm going to | |
| 02:12 | rewrite equation one here , you'll see why in a | |
| 02:15 | second , so two X plus four Y is equal | |
| 02:19 | to 10 . Now , what we're going to do | |
| 02:21 | is we're gonna multiply equation two by two , so | |
| 02:25 | two times X is equal to two X . This | |
| 02:28 | is going to be added to nine times two , | |
| 02:31 | which is 18 , That's in front of the variable | |
| 02:34 | y . And we have 12 times to which is | |
| 02:38 | 24 . Cool . Just double that entire equation . | |
| 02:41 | We can do that . So the next thing we | |
| 02:43 | do is we are going to pretty much eliminate one | |
| 02:46 | of the equations here . We have these matching coefficients | |
| 02:49 | and variables . We're now going to take one equation | |
| 02:53 | off the other and I'll show you how we do | |
| 02:54 | that . I look usually at what we have here | |
| 02:57 | is we look at this second variable of why the | |
| 03:00 | bigger one , I'm going to take away the smaller | |
| 03:03 | one here . So I'm going to multiply this entire | |
| 03:05 | equation I guess you could think of by negative one | |
| 03:08 | when I do that , I'm gonna end up with | |
| 03:10 | negative two X . This is going to become negative | |
| 03:14 | for why ? And this is going to equal negative | |
| 03:16 | 10 . Okay , now let's take one equation off | |
| 03:21 | the other by eliminating it . So what happens when | |
| 03:23 | we do this ? Well this is what we get | |
| 03:26 | to X minus two X . Well this is just | |
| 03:29 | going to give us nothing . So I'm not going | |
| 03:31 | to write anything down here 18 y minus four . | |
| 03:34 | Y . We're going to end up with 14 Y | |
| 03:37 | . And this is going to equal 24 minus 10 | |
| 03:41 | , 24 minus 10 is equal to 14 , so | |
| 03:44 | 14 y is equal to 14 . That means therefore | |
| 03:48 | y is equal to one because 14 divided by 14 | |
| 03:52 | is equal to one . So straightaway we have our | |
| 03:55 | first uh solving of one of our variables here , | |
| 03:59 | why is equal to one ? So I'm going to | |
| 04:01 | rub out all this here and we're just gonna remember | |
| 04:04 | that . Why is equal to one ? Obviously if | |
| 04:06 | you're in school you want to keep that all there | |
| 04:08 | so your teacher can see you're working out . But | |
| 04:11 | I have this uh little problem for space here . | |
| 04:14 | So we're gonna say that why is equal to one | |
| 04:16 | ? We work that one out just then . Now | |
| 04:18 | what we do is we substitute this value , Y | |
| 04:21 | equals one into one of these equations here to solve | |
| 04:24 | it . I think the easiest equation is equation to | |
| 04:27 | here . Okay , so let's put this into equation | |
| 04:30 | two , we have the X . We don't know | |
| 04:33 | that is , that's what we're trying to work out | |
| 04:35 | . And what we have is Y is equal to | |
| 04:37 | one , so nine times Y is equal to nine | |
| 04:40 | and this is equal to 12 . Pretty simple to | |
| 04:43 | solve . Now we could take nine off both sides | |
| 04:46 | and we end up with X is equal to three | |
| 04:48 | because three plus nine is equal to 12 . So | |
| 04:51 | we have our two values here , we have X | |
| 04:54 | is equal to three and y is equal to one | |
| 04:59 | . The next thing I do just as a little | |
| 05:00 | bit of a last thing here . That's good policy | |
| 05:03 | to now substitute in your X and Y value into | |
| 05:07 | the first equation or the other equation just to see | |
| 05:09 | if you got it correct . So let's do that | |
| 05:11 | . So two times X is equal to six , | |
| 05:14 | Okay , four times Y y is equal to one | |
| 05:17 | , so that is equal to 46 plus four is | |
| 05:20 | equal to 10 . We've got the correct answer there | |
| 05:23 | . Everything's cool . And that's how you go solving | |
| 05:26 | this particular set of equations Using the elimination method , | |
| 05:29 | I'll tell you what , I'm gonna put a bit | |
| 05:30 | of a harder one up now . So for our | |
| 05:32 | second example , let's try this particular set of equations | |
| 05:36 | here . This one is going to be a little | |
| 05:37 | bit harder , so three x minus two , Y | |
| 05:41 | Is equal to 31 . And the next one we | |
| 05:46 | have is two X . And that's going to be | |
| 05:49 | having three white added to it , and that's going | |
| 05:51 | to be equal to negative one . So straight away | |
| 05:54 | you can see a couple of complications possibly here . | |
| 05:57 | The first thing you might notice is we don't have | |
| 05:59 | a single thing that we could multiply other equation by | |
| 06:02 | to get the variable so they match . So we're | |
| 06:04 | going to deal with that . The second , the | |
| 06:05 | next thing negatives watch out for these guys when you're | |
| 06:08 | doing these particular types of equations , they are a | |
| 06:11 | really , really easy way of making mistakes . So | |
| 06:14 | the first thing I'm going to do is I'm going | |
| 06:15 | to label each of those equations . We have equation | |
| 06:17 | one and we have equation too . Now , what | |
| 06:21 | we're going to do is we're going to look what | |
| 06:23 | we can multiply an equation by to get the matching | |
| 06:27 | coefficients in front of the variable here . So safer | |
| 06:29 | X . What can we multiply two X by here | |
| 06:33 | to get three X . And I can't think of | |
| 06:36 | anything . Well it's going to be 1.5 . It's | |
| 06:37 | gonna get messy . What we're going to do is | |
| 06:39 | we're going to multiply each of the equations by a | |
| 06:42 | different number so we can get matching coefficients for the | |
| 06:46 | variables here . So As you have a look here | |
| 06:48 | , three and two and number the both of these | |
| 06:50 | guys go into is six to get to six , | |
| 06:54 | we would actually multiply three by two and to get | |
| 06:57 | to six we would multiply two by three . So | |
| 07:01 | that's what we're going to do to each of these | |
| 07:02 | equations here . So , equation one , let's multiply | |
| 07:05 | by two threats . Times two is equal to six | |
| 07:09 | X negative two Y times two is negative for why | |
| 07:15 | ? This is equal to 31 times two which is | |
| 07:19 | 62 . Watch out for those negatives and positives . | |
| 07:22 | Right . Let's put the two down there . The | |
| 07:24 | second equation . So two X times three is equal | |
| 07:29 | to six X matching just like we'd hoped . Uh | |
| 07:33 | positive three y times three is positive nine y at | |
| 07:38 | this is equal to negative one times three . This | |
| 07:42 | is going to be negative three . All right . | |
| 07:45 | So , we have these matching coefficients and variables here | |
| 07:47 | for X . So , let's now finish off this | |
| 07:50 | particular equation here . Now , the next step we | |
| 07:53 | do is we're going to end up taking one equation | |
| 07:56 | off the other , the elimination part . So we | |
| 07:58 | do that . Okay , to look here about which | |
| 08:00 | is my bigger one . This is the bigger one | |
| 08:02 | . So I'm gonna multiply this equation here by negative | |
| 08:06 | one . So multiplied by negative one . It's going | |
| 08:09 | to change all the sides . Be pretty methodical when | |
| 08:11 | you do this and watch out . It's very easy | |
| 08:13 | to make mistakes . So this becomes negative six X | |
| 08:16 | . This a negative times a negative . We're going | |
| 08:18 | to have positive for Y And 62 times -1 is | |
| 08:23 | -62 . All right , now let's solve this six | |
| 08:28 | X minus six X . That's where we're going to | |
| 08:31 | eliminate here . So , these guys are going to | |
| 08:32 | get rid of each other nine . Y plus four | |
| 08:36 | Y . Is equal to 13 Y . Okay , | |
| 08:41 | -3 -62 is equal to -65 . All right , | |
| 08:47 | so what have we got here ? 13 Why is | |
| 08:49 | equal to negative 65 . Okay , let's solve this | |
| 08:53 | . So we'll go negative 65 divided by 13 . | |
| 08:56 | We're going to get our answer of why equals negative | |
| 09:00 | five . Okay ? So we know that Y equals | |
| 09:03 | negative five . Now , let's go in and substitute | |
| 09:06 | this into one of our equations and see what we | |
| 09:09 | get . So let's do that right now . We | |
| 09:12 | know that why equals negative five . So , I'll | |
| 09:16 | get rid of these other guys as well . All | |
| 09:18 | right . Let's substitute in . So let's substitute into | |
| 09:21 | any one of them . I don't really mind . | |
| 09:23 | You don't So let's substitute into the first one here | |
| 09:25 | is as good as any into equation one . We | |
| 09:28 | have three X . And let's substitute negative five into | |
| 09:32 | this . A negative five times negative two is plus | |
| 09:36 | 10 . Watch out for your negatives and this is | |
| 09:38 | equal to 31 . All right , so what do | |
| 09:41 | we get now ? Three X is going to equal | |
| 09:44 | let's take 10 off this side and then 10 off | |
| 09:46 | this side is equal to 21 therefore X is equal | |
| 09:51 | to 21 divided by three , which is equal to | |
| 09:54 | seven . We've got both our answers , we got | |
| 09:56 | Y equals negative five and we have X is equal | |
| 09:59 | to seven . Let's just check this out in equation | |
| 10:02 | to hear . So X is equal to 72 times | |
| 10:05 | seven is equal to 14 . 3 times negative five | |
| 10:09 | is minus 15 . 14 minus 15 is equal to | |
| 10:13 | negative one . We have the correct answer . A | |
| 10:16 | really , really good thing to keep checking . Okay | |
| 10:19 | , what about one last one of these that you | |
| 10:21 | can do ? Okay , what about we do this | |
| 10:22 | one here , This is going to be six x | |
| 10:25 | minus three Y . And that is going to equal | |
| 10:29 | three And we're going to have four x . And | |
| 10:33 | this is Plus five Y . And that is going | |
| 10:37 | to equal 16 . All right , let's go through | |
| 10:40 | and solve it . So , the very first thing | |
| 10:42 | we do , you know , we're just going to | |
| 10:43 | give each one of these a label . We have | |
| 10:47 | equation one and equation to . You're going to see | |
| 10:49 | that we don't actually have anything that lines up nicely | |
| 10:52 | here . So let's look for one that we can | |
| 10:54 | do it with . Let's uh let's get let's look | |
| 10:56 | some variables that we can match up . We have | |
| 10:58 | six X . And we have for ex CIA . | |
| 11:02 | So a number of both of these going to , | |
| 11:04 | that's the simplest one I can think of is 12 | |
| 11:06 | . We would multiply this by two and we would | |
| 11:09 | multiply this by three . If we're looking at the | |
| 11:12 | coefficients in front of X here . So let's go | |
| 11:15 | an equation one and we have equation to here , | |
| 11:18 | let's now solve it . We end up with 12 | |
| 11:20 | X . That's what we should get two times six | |
| 11:22 | . X . 12 X . Uh negative three Y | |
| 11:26 | times two is negative six Y . And that is | |
| 11:30 | equal to three times two , which is six . | |
| 11:32 | That's equation one , equation 23 times four . Xs | |
| 11:37 | 12 X . That is going to be fired by | |
| 11:41 | times three . So that's positive 15 Y . And | |
| 11:45 | that is equal to 16 times three , which is | |
| 11:48 | equal to 48 . Alright , cool . Right now | |
| 11:52 | , what are we going to do ? Well , | |
| 11:53 | we've got that matching party of the 12 X . | |
| 11:56 | Is they're going to end up eliminating each other out | |
| 11:58 | sort of thing . So I'm now going to take | |
| 12:01 | the smaller one of these off . So let's do | |
| 12:03 | that . This is going to be the one I'm | |
| 12:05 | going to multiply by negative one . So this is | |
| 12:08 | going to become negative . That's going to become positive | |
| 12:11 | and this here is going to become negative . Okay | |
| 12:14 | ? All right . It's a really good policy that | |
| 12:15 | you do this . I know it seems like it's | |
| 12:17 | not a major step , but it's a very important | |
| 12:19 | one . So 12 X . Take away 12 X | |
| 12:22 | . They cancel each other out . 15 Y plus | |
| 12:26 | 15 Y is 21 Y . And this is equal | |
| 12:29 | to 48 . Subtract six , which is equal to | |
| 12:32 | 42 . You can see a really , really simple | |
| 12:35 | answer here we have that Y is equal to 42 | |
| 12:40 | divided by 21 which is equal to two . So | |
| 12:43 | that's the first part of our answer there we have | |
| 12:45 | Y is equal to two , and I'll move that | |
| 12:49 | up there and then we will substitute our values in | |
| 12:53 | and finish off this particular set of equations here . | |
| 12:57 | Okay , so let's do that now let's go through | |
| 13:00 | equation one , I guess equation one here . Unless | |
| 13:04 | substitute in our value for why here . So what | |
| 13:07 | do we have ? We have six X . That's | |
| 13:09 | going to stay the same negative three Y times two | |
| 13:13 | is going to be negative three times two is going | |
| 13:16 | to be minus six and that is equal two straight | |
| 13:20 | . All right . So what happens when we get | |
| 13:21 | this lets the solvent six X . And we're going | |
| 13:24 | to add six to both sides . This is equal | |
| 13:26 | to nine . Therefore X is going to equal nine | |
| 13:31 | divided by six , which is going to be 1.5 | |
| 13:35 | . Okay , bit tricky there . Right ? Not | |
| 13:37 | a whole number . So now it's just substitute into | |
| 13:40 | the second equation and check our answers are correct . | |
| 13:42 | We have X is equal to 1.5 , so four | |
| 13:45 | times 1.5 , Well double 1.5 is three and double | |
| 13:49 | again is equal to six and five times . Why | |
| 13:52 | ? Five times two is equal to 10 . 6 | |
| 13:55 | plus 10 is equal to 16 . Therefore our answer | |
| 13:59 | is correct anyway . Hopefully you like that video on | |
| 14:02 | the elimination method of simultaneous equations . If you did | |
| 14:06 | let me know in the comments and give me a | |
| 14:07 | thumbs up would be much appreciated . Anyway , we'll | |
| 14:10 | see you next time . Stay safe . Stay well | |
| 14:13 | , see you later . |
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