WEBVTT
00:00:01.600 --> 00:00:08.390
Here in this video, we’re gonna use the process of algebraic substitution to solve some linear simultaneous equations.
00:00:09.620 --> 00:00:13.910
It’s a process that isn’t always the most suitable method, but it can be really useful.
00:00:14.220 --> 00:00:19.610
So we’ll see some examples of when it works well and one or two where it actually makes life harder for us.
00:00:20.990 --> 00:00:31.240
Now remember that every point on a straight line graph represents a unique and different solution to the equation that represents that line.
00:00:32.820 --> 00:00:36.270
So for example, this point here I’ve got an 𝑥-value.
00:00:36.270 --> 00:00:42.340
So if I put an 𝑥-value of four into that equation, the corresponding 𝑦-value will be one.
00:00:44.500 --> 00:00:51.420
If I put an 𝑥-value of seven into the equation, the corresponding 𝑦-value that makes that true will be zero.
00:00:52.720 --> 00:01:02.760
And as we said every single point on that line represents a unique combination of 𝑥- and 𝑦-values that when I do the 𝑥-value and I add three times the 𝑦-value, I’m gonna get the answer seven.
00:01:03.350 --> 00:01:13.820
And for the other line 𝑥 plus 𝑦 equals three, again every point on that line represents another unique combination of 𝑥- and 𝑦-values that will sum to three.
00:01:15.500 --> 00:01:20.160
So 𝑥 is negative one, 𝑦 is four; they sum to three.
00:01:20.640 --> 00:01:25.100
𝑥 is two, 𝑦 is one; they sum to three and so on.
00:01:25.660 --> 00:01:36.640
But what’s so special about simultaneous equations is there in this particular case when we’ve got two lines, there is one pair of 𝑥- and 𝑦-values that give- that match both of those equations.
00:01:36.770 --> 00:01:38.690
They give three when you add them together.
00:01:38.960 --> 00:01:41.660
But if you add three times the 𝑦-coordinate to the 𝑥-coordinate.
00:01:41.660 --> 00:01:42.890
They also give seven.
00:01:43.050 --> 00:01:47.830
That’s this point here on the graph where the two lines intersect.
00:01:48.110 --> 00:01:52.760
So solving simultaneous equations is all about finding where those lines intersect.
00:01:53.970 --> 00:01:54.790
So here’s a question.
00:01:54.970 --> 00:02:03.020
Use algebraic substitution to solve the simultaneous equations 𝑦 equals three 𝑥 minus two and 𝑥 equals three 𝑦 minus ten.
00:02:03.760 --> 00:02:09.590
So they’ve told us the specific method we’ve got to use and they’ve told us that these two equations are simultaneous.
00:02:10.490 --> 00:02:17.120
Now it’s always good when you do number your equations so you can refer to those numbers when you’re explaining your working out.
00:02:17.390 --> 00:02:22.950
But I like to put a little brace to indicate that those two things are simultaneously true as well.
00:02:23.000 --> 00:02:29.920
And not everybody does that, but it’s I just find it’s a good idea to communicate this idea or simultaneousness of the equations.
00:02:31.170 --> 00:02:38.840
Now the thing about substitution is in the first equation, we’ve got the fact that 𝑦 is equal to this bunch of stuff here three 𝑥 plus [minus] two.
00:02:39.330 --> 00:02:48.510
So substitution is all about saying- well in the second equation, we’ll replace 𝑦 with all that bunch of stuff because those two things are true at the same time.
00:02:48.510 --> 00:02:55.570
So whilst 𝑦 is represented in the second equation, we also want it to be equal to three 𝑥 minus two.
00:02:56.950 --> 00:03:04.200
So in this case we’re gonna take the value that 𝑦 has in equation one and substitute 𝑦 in the second equation.
00:03:05.640 --> 00:03:07.910
And we could have done that the other way around if we wanted to.
00:03:08.300 --> 00:03:11.390
Equation two says that 𝑥 is equal to three 𝑦 minus ten.
00:03:11.390 --> 00:03:15.270
So we could’ve replaced 𝑥 in the first equation with three 𝑦 minus ten.
00:03:15.300 --> 00:03:20.400
So it doesn’t matter which way around you do it as long as you’re completely replacing the letter that you’re trying to get rid of.
00:03:20.950 --> 00:03:22.490
So now we’ve got an equation here.
00:03:22.610 --> 00:03:27.960
Let’s call that equation number three, which is purely in terms of 𝑥.
00:03:27.990 --> 00:03:31.110
So we’re gonna be able to find a unique solution to this for 𝑥.
00:03:32.360 --> 00:03:42.290
So using the distributive law of multiplication, 𝑥 is equal to three lots of three 𝑥; so that’s nine 𝑥 and three times negative two which is negative six.
00:03:42.720 --> 00:03:44.510
And then we’ve got to take away ten as well.
00:03:45.730 --> 00:03:52.060
So 𝑥 is equal to nine 𝑥 minus six minus another ten; that’s nine 𝑥 minus sixteen.
00:03:52.510 --> 00:03:54.740
Now I want to get all the 𝑥s onto one side.
00:03:54.740 --> 00:03:58.380
And it’s generally speaking a good idea to have a positive number of 𝑥s when we do that.
00:03:58.600 --> 00:04:04.710
So I’m gonna just subtract one 𝑥 from both sides leaving me with eight 𝑥 on the right-hand side and no 𝑥s on the left-hand side.
00:04:05.650 --> 00:04:10.740
So we’re subtracting 𝑥 on both sides and as we said 𝑥 minus 𝑥 on the left-hand side, it gives us zero.
00:04:11.920 --> 00:04:15.840
And nine 𝑥 take away 𝑥 on the right-hand side leaves us with eight 𝑥.
00:04:16.940 --> 00:04:20.940
Now I can add sixteen to both sides to get rid of that number from the right-hand side.
00:04:21.750 --> 00:04:25.420
And on the left-hand side, zero plus sixteen is just sixteen.
00:04:26.170 --> 00:04:29.470
And on the right-hand side, negative sixteen plus sixteen is zero.
00:04:30.570 --> 00:04:33.160
So we’re left with sixteen is equal to eight 𝑥.
00:04:33.550 --> 00:04:35.240
Now I wanna know what one 𝑥 is.
00:04:35.240 --> 00:04:41.260
So if I divide both sides by eight, on the left-hand side sixteen divided by eight is two.
00:04:42.280 --> 00:04:46.300
And on the right-hand side the eights cancel out to leave me with just 𝑥.
00:04:47.550 --> 00:04:50.600
So I know that 𝑥 is equal to two.
00:04:51.300 --> 00:04:53.050
Now I want to know what 𝑦 is equal to.
00:04:53.080 --> 00:04:57.950
But remember equation one has got an equation which is 𝑦 is equal to three 𝑥 minus two.
00:04:57.950 --> 00:05:03.170
So if I substitute that value of 𝑥 back into equation one, it’ll tell me straight away what 𝑦 is.
00:05:04.350 --> 00:05:07.140
So 𝑦 is three times two take away two.
00:05:08.060 --> 00:05:12.170
So that’s six take away two, which means that 𝑦 is four.
00:05:13.380 --> 00:05:14.950
Now we can check our answers.
00:05:15.180 --> 00:05:20.450
We just used substitution in equation one to work out that 𝑦 was equal to four.
00:05:20.450 --> 00:05:24.690
So I’m gonna use the other equation — equation two — just to check that my 𝑥 and 𝑦 match up.
00:05:25.930 --> 00:05:29.840
So I’ve substituted in two for 𝑥 and four for 𝑦.
00:05:30.150 --> 00:05:33.540
And I’ve got two is equal to three times four take away ten.
00:05:35.020 --> 00:05:38.110
So two is equal to twelve minus ten which is two.
00:05:38.110 --> 00:05:40.260
That’s correct; so we’re happy with our answer.
00:05:41.580 --> 00:05:45.280
And so we can just put a nice box around our answer to make it lovely and clear.
00:05:46.240 --> 00:05:52.730
So that’s an example where you’ve got a ready-made equation that says 𝑦 is equal to something; 𝑥 is equal to something.
00:05:52.990 --> 00:05:57.450
And it’s very easy to substitute either 𝑥 or 𝑦 into the other equation.
00:05:58.710 --> 00:06:00.310
So let’s move on to a second example.
00:06:00.340 --> 00:06:08.950
Use algebraic substitution to solve the simultaneous equations 𝑥 is equal to two 𝑦 plus eight and two 𝑥 plus three 𝑦 equals twenty-three.
00:06:09.810 --> 00:06:16.460
Now looking at that first equation, we’ve got 𝑥 is equal to something not involving 𝑥, so two 𝑦 plus eight.
00:06:16.780 --> 00:06:20.250
So there’s an obvious thing that we can use to substitute into the second equation.
00:06:20.710 --> 00:06:30.500
In order to find out what 𝑦 is equal to in the second equation or even what 𝑥 is equal to to substitute it back into the first, we’d have to do quite a bit of work to rearrange it.
00:06:30.940 --> 00:06:37.170
So the obvious substitution is to take 𝑥 from equation one and substitute it into equation two.
00:06:38.670 --> 00:06:43.800
So substitution is just a matter of saying in the first equation 𝑥 is equal to all this stuff here.
00:06:44.250 --> 00:06:48.940
So where we see 𝑥 in our second equation, we’ll replace 𝑥 with all that stuff.
00:06:50.270 --> 00:06:57.820
Now we’re going to multiply out the parentheses and solve that equation to find out the value of 𝑦.
00:06:58.400 --> 00:07:01.750
So two lots of two 𝑦 is four 𝑦; two lots of eight equals sixteen.
00:07:01.750 --> 00:07:04.850
We’ve still got our three 𝑦 and that’s all equal to twenty-three.
00:07:05.500 --> 00:07:08.840
Now we’ve got four 𝑦 and three 𝑦; so that’s seven 𝑦.
00:07:10.100 --> 00:07:12.630
So that seven 𝑦 plus sixteen equals twenty-three.
00:07:12.660 --> 00:07:17.270
Now if I subtract sixteen from both sides, I’ll be just left with seven 𝑦 on the left.
00:07:18.620 --> 00:07:21.010
So seven 𝑦 plus sixteen take away sixteen.
00:07:21.010 --> 00:07:24.310
Well sixteen take away sixteen is nothing; so that wipes that out.
00:07:24.310 --> 00:07:28.560
We’ve just got seven 𝑦 and twenty-three take away sixteen is just seven.
00:07:29.610 --> 00:07:31.610
So seven 𝑦 is equal to seven.
00:07:31.960 --> 00:07:38.170
Now if I divide both sides by seven, I’ll find that 𝑦 is equal to one.
00:07:39.570 --> 00:07:43.840
And luckily our first equation told us the value of 𝑥 in terms of 𝑦.
00:07:43.840 --> 00:07:49.290
So I just need to substitute that value of 𝑦 back into that equation and I’ll find out what 𝑥 is.
00:07:50.440 --> 00:07:57.860
So if 𝑥 was two times 𝑦 plus eight and we know that 𝑦 is equal to one, that means that 𝑥 is equal to two times one plus eight.
00:07:57.970 --> 00:07:58.860
So 𝑥 is ten.
00:08:00.070 --> 00:08:08.370
Always a good idea to check your answers, so we’ll use the other equation, equation two, just to substitute 𝑥 is ten and 𝑦 is one and just check that it all works out.
00:08:09.840 --> 00:08:18.070
So if all is good, we’ve got two times 𝑥 is ten plus three times 𝑦 is one and that will be equal to twenty-three.
00:08:19.250 --> 00:08:21.760
Well two times ten is twenty and three times one is three.
00:08:21.760 --> 00:08:24.500
So yep that’s true; looks like we’ve got the right answer.
00:08:25.630 --> 00:08:27.950
So that’s pretty much the substitution method.
00:08:30.230 --> 00:08:31.780
So let’s look at number three then.
00:08:31.810 --> 00:08:40.810
Use algebraic substitution to solve the simultaneous equations three 𝑥 plus three 𝑦 equals twenty-seven and two 𝑥 plus five 𝑦 equals thirty-six.
00:08:41.950 --> 00:08:50.810
Now just looking at this-this pair of equations, I- my gut reaction would be — if they didn’t tell me which method to use I’d probably be leaning towards elimination.
00:08:51.060 --> 00:09:02.030
I’d be perhaps doing two times equation one and three times equation two and then subtracting one from the other and eliminating 𝑥 from our enquiries, working out what 𝑦 is and then substituting that back in.
00:09:02.450 --> 00:09:06.270
But that’s not the case; we’ve been told specifically we’ve gotta use algebraic substitution.
00:09:06.270 --> 00:09:13.470
So we need to rearrange one of those equations to get either 𝑥 or 𝑦 on its own and then substitute that into the other equation.
00:09:15.420 --> 00:09:23.370
Now looking at both of those equations, I think I’m gonna mess about with equation one because the coefficients of 𝑥 and 𝑦 are both three.
00:09:23.370 --> 00:09:25.040
And twenty-seven is a multiple of three.
00:09:25.040 --> 00:09:29.960
So if I get 𝑥 or 𝑦 on its own, I can divide everything through by three.
00:09:29.960 --> 00:09:32.480
And I won’t have any fractions involved in this day.
00:09:32.480 --> 00:09:33.400
So let’s try that.
00:09:34.750 --> 00:09:37.270
So I’m gonna try to make 𝑦 the subject.
00:09:37.520 --> 00:09:40.950
So first of all I need to get rid of the three 𝑥 term.
00:09:40.950 --> 00:09:49.130
So I’m gonna subtract three 𝑥 from both sides of the equation, so subtracting three 𝑥 from this side and subtracting three 𝑥 from this side.
00:09:50.500 --> 00:09:57.840
Well subtracting three 𝑥 from the left-hand side just leaves me with three 𝑦 and subtracting three 𝑥 from the right-hand side.
00:09:58.560 --> 00:10:01.260
I’m gonna write negative three 𝑥 plus twenty-seven.
00:10:01.290 --> 00:10:03.360
You could say twenty-seven minus three 𝑥.
00:10:03.360 --> 00:10:06.690
It wouldn’t make any difference in the long run, but I’m just gonna do this for now.
00:10:07.350 --> 00:10:09.100
Now that’s what three 𝑦 is.
00:10:09.850 --> 00:10:12.900
So I need to divide everything by three in order to get what one 𝑦 is.
00:10:14.160 --> 00:10:23.450
So a third of three 𝑦 is just 𝑦, a third of negative three 𝑥 is just negative 𝑥, and a third of twenty-seven is positive nine.
00:10:24.840 --> 00:10:30.740
So from equation one, we know that 𝑦 is equal to negative 𝑥 plus nine; it’s equal to all that stuff.
00:10:31.080 --> 00:10:36.360
Now we can substitute that version of 𝑦 negative 𝑥 plus nine into the second equation.
00:10:37.420 --> 00:10:41.710
So remember our second equation was two 𝑥 plus five 𝑦 equals thirty-six.
00:10:41.980 --> 00:10:45.890
So two 𝑥 plus five 𝑦; now we’re saying 𝑦 is equal to all this stuff.
00:10:46.140 --> 00:10:50.820
So we’re gonna put this stuff instead of 𝑦 five times that and that’s equal to thirty-six.
00:10:50.950 --> 00:11:03.700
So now we’re gonna multiply out the parentheses here, which gives us two 𝑥 minus five 𝑥 plus forty-five equals thirty-six and two 𝑥 minus five 𝑥 is negative three 𝑥.
00:11:04.510 --> 00:11:09.650
So I would advise adding three 𝑥 to both sides to give me a positive number of 𝑥s somewhere.
00:11:10.900 --> 00:11:14.650
So adding three 𝑥 to the left-hand side just leaves me with forty-five.
00:11:15.040 --> 00:11:19.770
And adding three 𝑥 to the right-hand side, I can say three 𝑥 plus thirty-six or thirty-six plus three 𝑥.
00:11:19.770 --> 00:11:22.520
I’m gonna do it this way round this time just for a change.
00:11:23.990 --> 00:11:29.590
And then I’m gonna subtract thirty-six from both sides, just to leave the 𝑥 term on its own on the right-hand side.
00:11:31.120 --> 00:11:35.980
So subtracting thirty-six on the right-hand side obviously just leaves me with three 𝑥.
00:11:36.270 --> 00:11:41.130
And subtracting thirty-six on the left-hand side, so forty-five minus thirty-six is nine.
00:11:41.640 --> 00:11:43.340
So three 𝑥 is equal to nine.
00:11:43.470 --> 00:11:47.370
So now I can divide both sides by three to tell me what one 𝑥 is.
00:11:48.970 --> 00:11:53.070
Third of three 𝑥 is one 𝑥 and third of nine is three; so 𝑥 is equal to three.
00:11:54.460 --> 00:11:58.960
Now down here we had said that 𝑦 is equal to negative 𝑥 plus nine.
00:11:58.960 --> 00:12:01.110
Let’s call that equation three, shall we?
00:12:01.490 --> 00:12:06.210
So using the equation number three, 𝑦 is equal to negative 𝑥 plus nine.
00:12:06.210 --> 00:12:10.260
We can now substitute 𝑥 is three into that equation to find out what 𝑦 is.
00:12:11.460 --> 00:12:18.550
So 𝑦 is the negative of the 𝑥-value; so negative of three plus nine which is equal to six.
00:12:19.720 --> 00:12:24.750
And now I’m gonna check those values back in one of my original equations just to see that everything adds up properly.
00:12:24.750 --> 00:12:26.870
So I’m gonna go for equation two.
00:12:26.900 --> 00:12:28.000
That looks more interesting.
00:12:29.520 --> 00:12:35.370
And equation two said two times the 𝑥-value plus five times the 𝑦-value equals thirty-six.
00:12:35.370 --> 00:12:40.220
That’s two times three plus five times six, which is six plus thirty.
00:12:40.220 --> 00:12:41.770
And yup that is thirty-six.
00:12:43.070 --> 00:12:46.420
So our answer is 𝑥 is three and 𝑦 is six.
00:12:48.200 --> 00:13:01.610
So when we had you know complicated equations at the beginning that didn’t have a nice simple 𝑦 equals or 𝑥 equals that we can substitute into the other one, then we had quite a bit of work to do before we could actually get on to do the substitution.
00:13:01.900 --> 00:13:04.180
So just bear that in mind when using this method.
00:13:05.970 --> 00:13:16.230
Now number four, we’ve got to use algebraic substitution to solve the simultaneous equations four 𝑥 plus three 𝑦 equals three and five 𝑥 plus four 𝑦 equals three and eleven-twelfths.
00:13:17.720 --> 00:13:28.160
So not-not only have we not got an obvious 𝑦 equals or 𝑥 equals, but there’s not even a very nice way to rearrange these to get nice numbers without fractions in them.
00:13:28.400 --> 00:13:36.170
So you know if I wanted to make 𝑦 the subject in either of those, I’m gonna have some fraction of 𝑥 and then fractional numbers; it’s all very horrible.
00:13:37.920 --> 00:13:43.730
So this is probably a classic example of where you would not use algebraic substitution to solve the simultaneous equation.
00:13:44.120 --> 00:13:56.130
I’m gonna show you it quickly anyway so that you can see how horrible it turns and just so that you can avoid it in future yourself if you come across ones like these and if they don’t tell you that you’ve got to use algebraic substitution that is.
00:13:57.430 --> 00:14:01.090
So I’m gonna rearrange equation one to make 𝑦 the subject.
00:14:01.090 --> 00:14:07.920
So I’ve taken away four 𝑥 from both sides giving me three 𝑦 is three minus four 𝑥 and then divided every term on both sides by three.
00:14:07.920 --> 00:14:13.980
So I’ve got 𝑦 is equal to this stuff here: one minus four-thirds of 𝑥.
00:14:14.300 --> 00:14:15.770
And I’ve called that equation three.
00:14:16.800 --> 00:14:22.520
So I’m gonna substitute that version, that value of 𝑦, into equation two.
00:14:23.790 --> 00:14:29.060
So we’ve taken that value of 𝑦 and we’ve replaced 𝑦 in the equation with that value.
00:14:29.530 --> 00:14:35.990
The other thing that I’ve done here is I converted the mixed number three and eleven-twelfths into forty-seven over twelve.
00:14:35.990 --> 00:14:41.770
It’s generally easier to do your calculations with top heavy fractions rather than messing about with mixed numbers.
00:14:42.720 --> 00:14:48.400
So now we’re gonna multiply out the brackets four times one and four times negative four-thirds 𝑥.
00:14:50.020 --> 00:14:56.190
And that gives us a five 𝑥 plus four minus sixteen-thirds 𝑥 equals forty-seven over twelve.
00:14:56.190 --> 00:14:59.370
So we’ve got five 𝑥 and we’re gonna take away sixteen-thirds of 𝑥.
00:14:59.930 --> 00:15:08.910
So really we want to express five 𝑥 in a top heavy fraction with a denominator of three to end this easiest; so fifteen over three 𝑥.
00:15:10.210 --> 00:15:12.780
So five 𝑥 is the same as fifteen over three 𝑥.
00:15:12.780 --> 00:15:16.680
We’ve now got fifteen over three 𝑥 subtract sixteen over three 𝑥.
00:15:16.710 --> 00:15:21.220
So that’s negative one over three 𝑥, so negative a third 𝑥.
00:15:21.220 --> 00:15:26.190
If I add a third 𝑥 to both sides, then I’m gonna end up with a positive number of 𝑥s over on the right-hand side.
00:15:27.250 --> 00:15:30.630
And then I’m gonna subtract forty-seven over twelve from both sides.
00:15:30.920 --> 00:15:35.830
But at the same time, I’m gonna convert four into a top heavy fraction involving twelve.
00:15:35.830 --> 00:15:38.910
So that’s gonna be forty-eight over twelve, which is the same as four.
00:15:40.390 --> 00:15:45.780
So four becomes forty-eight over twelve with subtracting forty-seven over twelve.
00:15:47.710 --> 00:15:53.630
And when I subtract forty-seven over twelve from the right-hand side, it gets rid of that; it just leaves me with a third of 𝑥.
00:15:54.070 --> 00:15:58.680
So forty-eight-twelfths minus forty-seven-twelfths is just one-twelfth.
00:15:59.070 --> 00:16:03.680
Now to find out what 𝑥 is equal to, I’m gonna multiply both sides by three.
00:16:04.950 --> 00:16:10.450
So 𝑥 is equal to three-twelfths, which is obviously the same as a quarter.
00:16:11.870 --> 00:16:20.980
Now I can substitute that value of 𝑥 back into this equation we had up here: 𝑦 equals one minus four-thirds of 𝑥 and that will tell me what 𝑦 is equal to.
00:16:22.360 --> 00:16:26.250
So 𝑦 is one minus four-thirds times a quarter.
00:16:26.620 --> 00:16:31.010
Well the four is gonna cancel here; so that’s one-third.
00:16:31.010 --> 00:16:34.880
So 𝑦 is equal to one minus a third which is two-thirds.
00:16:35.820 --> 00:16:40.720
Now I’ll leave it to you to go back and check that in one of the original equations.
00:16:41.120 --> 00:16:51.090
But take it from me once you start doing algebraic substitution on unsuitable pairs of equations, it does get quite tricky; there’s lots of fractions and lots of negative numbers.
00:16:51.090 --> 00:16:52.180
So it all gets a bit messy.
00:16:54.160 --> 00:16:59.620
Yep so overall the algebraic substitution method is okay in some cases for linear equations.
00:16:59.620 --> 00:17:04.330
But it really comes into its own when you’ve got a nonlinear equation and a linear equation.
00:17:04.330 --> 00:17:06.230
And you can do a nice simple substitution that way.
00:17:06.230 --> 00:17:12.500
So why not check out that particular video solving simultaneous nonlinear equations using algebraic substitution.
00:17:13.310 --> 00:17:14.220
Okay thanks for now.