Video Transcript
Here in this video, we’re gonna use
the process of algebraic substitution to solve some linear simultaneous
equations. It’s a process that isn’t always
the most suitable method, but it can be really useful. So we’ll see some examples of when
it works well and one or two where it actually makes life harder for us.
Now remember that every point on a
straight line graph represents a unique and different solution to the equation that
represents that line. So for example, this point here
I’ve got an 𝑥-value. So if I put an 𝑥-value of four
into that equation, the corresponding 𝑦-value will be one. If I put an 𝑥-value of seven into
the equation, the corresponding 𝑦-value that makes that true will be zero. 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. 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.
So 𝑥 is negative one, 𝑦 is four;
they sum to three. 𝑥 is two, 𝑦 is one; they sum to
three and so on. 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. They give three when you add them
together. But if you add three times the
𝑦-coordinate to the 𝑥-coordinate. They also give seven. That’s this point here on the graph
where the two lines intersect. So solving simultaneous equations
is all about finding where those lines intersect.
So here’s a question.
Use algebraic substitution to
solve the simultaneous equations 𝑦 equals three 𝑥 minus two and 𝑥 equals
three 𝑦 minus ten. So they’ve told us the specific
method we’ve got to use and they’ve told us that these two equations are
simultaneous. 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. But I like to put a little
brace to indicate that those two things are simultaneously true as well. 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. 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. 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. So whilst 𝑦 is represented in
the second equation, we also want it to be equal to three 𝑥 minus two. So in this case we’re gonna
take the value that 𝑦 has in equation one and substitute 𝑦 in the second
equation.
And we could have done that the
other way around if we wanted to. Equation two says that 𝑥 is
equal to three 𝑦 minus ten. So we could’ve replaced 𝑥 in
the first equation with three 𝑦 minus ten. 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. So now we’ve got an equation
here. Let’s call that equation number
three, which is purely in terms of 𝑥. So we’re gonna be able to find
a unique solution to this for 𝑥. 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. And then we’ve got to take away
ten as well. So 𝑥 is equal to nine 𝑥 minus
six minus another ten; that’s nine 𝑥 minus sixteen. Now I want to get all the 𝑥s
onto one side. And it’s generally speaking a
good idea to have a positive number of 𝑥s when we do that. 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. So we’re subtracting 𝑥 on both
sides and as we said 𝑥 minus 𝑥 on the left-hand side, it gives us zero. And nine 𝑥 take away 𝑥 on the
right-hand side leaves us with eight 𝑥. Now I can add sixteen to both
sides to get rid of that number from the right-hand side. And on the left-hand side, zero
plus sixteen is just sixteen. And on the right-hand side,
negative sixteen plus sixteen is zero. So we’re left with sixteen is
equal to eight 𝑥. Now I wanna know what one 𝑥
is. So if I divide both sides by
eight, on the left-hand side sixteen divided by eight is two. And on the right-hand side the
eights cancel out to leave me with just 𝑥. So I know that 𝑥 is equal to
two. Now I want to know what 𝑦 is
equal to. But remember equation one has
got an equation which is 𝑦 is equal to three 𝑥 minus two. So if I substitute that value
of 𝑥 back into equation one, it’ll tell me straight away what 𝑦 is. So 𝑦 is three times two take
away two. So that’s six take away two,
which means that 𝑦 is four.
Now we can check our
answers. We just used substitution in
equation one to work out that 𝑦 was equal to four. So I’m gonna use the other
equation — equation two — just to check that my 𝑥 and 𝑦 match up. So I’ve substituted in two for
𝑥 and four for 𝑦. And I’ve got two is equal to
three times four take away ten. So two is equal to twelve minus
ten which is two. That’s correct; so we’re happy
with our answer. And so we can just put a nice
box around our answer to make it lovely and clear. 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. And it’s very easy to
substitute either 𝑥 or 𝑦 into the other equation.
So let’s move on to a second
example.
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. Now looking at that first
equation, we’ve got 𝑥 is equal to something not involving 𝑥, so two 𝑦 plus
eight. So there’s an obvious thing
that we can use to substitute into the second equation. 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. So the obvious substitution is
to take 𝑥 from equation one and substitute it into equation two.
So substitution is just a
matter of saying in the first equation 𝑥 is equal to all this stuff here. So where we see 𝑥 in our
second equation, we’ll replace 𝑥 with all that stuff. Now we’re going to multiply out
the parentheses and solve that equation to find out the value of 𝑦. So two lots of two 𝑦 is four
𝑦; two lots of eight equals sixteen. We’ve still got our three 𝑦
and that’s all equal to twenty-three. Now we’ve got four 𝑦 and three
𝑦; so that’s seven 𝑦. So that seven 𝑦 plus sixteen
equals twenty-three. Now if I subtract sixteen from
both sides, I’ll be just left with seven 𝑦 on the left. So seven 𝑦 plus sixteen take
away sixteen. Well sixteen take away sixteen
is nothing; so that wipes that out. We’ve just got seven 𝑦 and
twenty-three take away sixteen is just seven. So seven 𝑦 is equal to
seven. Now if I divide both sides by
seven, I’ll find that 𝑦 is equal to one.
And luckily our first equation
told us the value of 𝑥 in terms of 𝑦. So I just need to substitute
that value of 𝑦 back into that equation and I’ll find out what 𝑥 is. 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. So 𝑥 is ten. 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. 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. Well two times ten is twenty
and three times one is three. So yep that’s true; looks like
we’ve got the right answer. So that’s pretty much the
substitution method.
So let’s look at number three
then.
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. 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. 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. But that’s not the case; we’ve
been told specifically we’ve gotta use algebraic substitution. 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. 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. And twenty-seven is a multiple
of three. So if I get 𝑥 or 𝑦 on its
own, I can divide everything through by three. And I won’t have any fractions
involved in this day. So let’s try that.
So I’m gonna try to make 𝑦 the
subject. So first of all I need to get
rid of the three 𝑥 term. 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. Well subtracting three 𝑥 from
the left-hand side just leaves me with three 𝑦 and subtracting three 𝑥 from
the right-hand side. I’m gonna write negative three
𝑥 plus twenty-seven. You could say twenty-seven
minus three 𝑥. It wouldn’t make any difference
in the long run, but I’m just gonna do this for now. Now that’s what three 𝑦 is. So I need to divide everything by three in order to get what one 𝑦 is. 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.
So from equation one, we know
that 𝑦 is equal to negative 𝑥 plus nine; it’s equal to all that stuff. Now we can substitute that
version of 𝑦 negative 𝑥 plus nine into the second equation. So remember our second equation
was two 𝑥 plus five 𝑦 equals thirty-six. So two 𝑥 plus five 𝑦; now
we’re saying 𝑦 is equal to all this stuff. So we’re gonna put this stuff
instead of 𝑦 five times that and that’s equal to thirty-six. 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. So I would advise adding three
𝑥 to both sides to give me a positive number of 𝑥s somewhere. So adding three 𝑥 to the
left-hand side just leaves me with forty-five. And adding three 𝑥 to the
right-hand side, I can say three 𝑥 plus thirty-six or thirty-six plus three
𝑥. I’m gonna do it this way round
this time just for a change. 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. So subtracting thirty-six on
the right-hand side obviously just leaves me with three 𝑥. And subtracting thirty-six on
the left-hand side, so forty-five minus thirty-six is nine. So three 𝑥 is equal to
nine. So now I can divide both sides
by three to tell me what one 𝑥 is. Third of three 𝑥 is one 𝑥 and
third of nine is three; so 𝑥 is equal to three.
Now down here we had said that
𝑦 is equal to negative 𝑥 plus nine. Let’s call that equation three,
shall we? So using the equation number
three, 𝑦 is equal to negative 𝑥 plus nine. We can now substitute 𝑥 is
three into that equation to find out what 𝑦 is. So 𝑦 is the negative of the
𝑥-value; so negative of three plus nine which is equal to six. And now I’m gonna check those
values back in one of my original equations just to see that everything adds up
properly. So I’m gonna go for equation
two. That looks more
interesting. And equation two said two times
the 𝑥-value plus five times the 𝑦-value equals thirty-six. That’s two times three plus
five times six, which is six plus thirty. And yup that is thirty-six. So our answer is 𝑥 is three
and 𝑦 is six. 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. So just bear that in mind when
using this method.
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.
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. 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. So this is probably a classic
example of where you would not use algebraic substitution to solve the
simultaneous equation. 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.
So I’m gonna rearrange equation
one to make 𝑦 the subject. 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. So I’ve got 𝑦 is equal to this
stuff here: one minus four-thirds of 𝑥. And I’ve called that equation
three. So I’m gonna substitute that
version, that value of 𝑦, into equation two. So we’ve taken that value of 𝑦
and we’ve replaced 𝑦 in the equation with that value. The other thing that I’ve done
here is I converted the mixed number three and eleven-twelfths into forty-seven
over twelve. It’s generally easier to do
your calculations with top heavy fractions rather than messing about with mixed
numbers. So now we’re gonna multiply out
the brackets four times one and four times negative four-thirds 𝑥. And that gives us a five 𝑥
plus four minus sixteen-thirds 𝑥 equals forty-seven over twelve. So we’ve got five 𝑥 and we’re
gonna take away sixteen-thirds of 𝑥. 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. So five 𝑥 is the same as
fifteen over three 𝑥. We’ve now got fifteen over
three 𝑥 subtract sixteen over three 𝑥. So that’s negative one over
three 𝑥, so negative a third 𝑥. 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.
And then I’m gonna subtract
forty-seven over twelve from both sides. But at the same time, I’m gonna
convert four into a top heavy fraction involving twelve. So that’s gonna be forty-eight
over twelve, which is the same as four. So four becomes forty-eight
over twelve with subtracting forty-seven over twelve. 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. So forty-eight-twelfths minus
forty-seven-twelfths is just one-twelfth. Now to find out what 𝑥 is equal to,
I’m gonna multiply both sides by three. So 𝑥 is equal to
three-twelfths, which is obviously the same as a quarter. 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. So 𝑦 is one minus four-thirds
times a quarter. Well the four is gonna cancel
here; so that’s one-third. So 𝑦 is equal to one minus a
third which is two-thirds.
Now I’ll leave it to you to go back
and check that in one of the original equations. 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. So it all gets a bit messy. Yep so overall the algebraic
substitution method is okay in some cases for linear equations. But it really comes into its own
when you’ve got a nonlinear equation and a linear equation. And you can do a nice simple
substitution that way. So why not check out that
particular video solving simultaneous nonlinear equations using algebraic
substitution. Okay thanks for now.
