Video Transcript
In this video, we will learn how to
solve systems of linear equations using elimination.
Linear equations are equations in
which the highest power of each variable that appears is one. And we also don’t have any terms in
which different variables are multiplied together. For example, the equation two 𝑥
plus three 𝑦 equals seven is a linear equation. It involves two variables or
unknowns, 𝑥 and 𝑦. And as it involves two variables,
we can’t solve this equation on its own; we need more information.
If we were given a second linear
equation also involving these same two variables such as the equation five 𝑥 minus
𝑦 equals nine, then we now have what’s called a system of linear equations. In general, we need the same number
of equations as there are variables. So in this system, we have two
variables 𝑥 and 𝑦 and two equations. So we’d be able to solve it in
order to find the values of 𝑥 and 𝑦 that work in both equations. There are a range of methods that
we can use to do this. In this video, we’re going to focus
on the elimination method. Let’s look then at a series of
different examples, beginning with this one.
Using elimination, solve the
simultaneous equations three 𝑥 plus two 𝑦 equals 14, six 𝑥 minus two 𝑦
equals 22.
Now, simultaneous equations are
just another way of saying a system of equations. We can see that we have two
equations. And they’re each in the same
two variables, 𝑥 and 𝑦. We’re told that we need to use
the method of elimination to answer this question. So let’s see what this looks
like. The principle of this method is
that we can eliminate or get rid of one of the two variables from our two
equations. We can choose to eliminate
either 𝑥 or 𝑦. But to make things easier, we
notice that in this question, we have two 𝑦 in each equation. But in equation 1, two 𝑦 is
being added to the 𝑥-term and in equation 2 it’s being subtracted from the
𝑥-term.
The key thing we need to spot
is that if we were to add these two entire equations together, then we’ll
eliminate the 𝑦-term. Let’s see what that looks
like. On the left-hand side, three 𝑥
plus six 𝑥 gives nine 𝑥. We then have positive two 𝑦
plus negative two 𝑦. That’s two 𝑦 minus two 𝑦,
which is equal to zero. On the right-hand side, we have
14 plus 22, which is equal to 36. So we’ve eliminated the
𝑦-variables and created an equation in 𝑥 only. Nine 𝑥 is equal to 36.
Now that our equation is in
terms of 𝑥 only, it’s straightforward to solve to find the value of 𝑥. We have nine 𝑥 equals 36, so
we need to divide both sides of the equation by nine. Doing so gives 𝑥 equals
four. So we’ve found the value of one
of our two variables. Next, we need to find the value
of our 𝑦-variable. And to do this, we can
substitute the value we’ve found for 𝑥 into either of our two equations. It really doesn’t matter which
we choose. I’m going to choose to use
equation 1 simply because the coefficient of 𝑦 is positive in this equation, so
it will make things a little easier.
So substituting 𝑥 equals four
gives three times four plus two 𝑦 is equal to 14. Three times four is, of course,
equal to 12. So we have the equation 12 plus
two 𝑦 equals 14, which is an equation in 𝑦 only. To solve, we first need to
subtract 12 from each side to give two 𝑦 is equal to two and then divide each
side of the equation by two to give 𝑦 is equal to one. So we’ve also found the value
of 𝑦. And therefore, we’ve solved the
simultaneous equations. Our solution is a pair of
values: 𝑥 is equal to four and 𝑦 is equal to one.
Now it’s always a good idea to
check our answer where we can. And in order to do this, we’re
going to substitute the pair of values we found for 𝑥 and 𝑦 into whichever
equation we didn’t use when determining the second value. So we’re going to substitute
into equation 2. Substituting 𝑥 equals four and
𝑦 equals one on the left-hand side gives six multiplied by four minus two
multiplied by one. That’s 24 minus two, which is
equal to 22. And that is indeed the value
that we should have on the right-hand side of the equation. So this confirms that our
solution is correct.
The key principle of the method
of elimination in this question then was to notice that we had almost the same
coefficient of 𝑦 in each equation, but one was positive and one was
negative. We, therefore, found that if we
were to add the two equations together, this would eliminate the 𝑦-variables,
leaving an equation in 𝑥 only. Our solution is 𝑥 equals four
and 𝑦 equals one.
In our next example, we’ll see how
the method of elimination works if we can’t eliminate one variable by adding the two
equations together.
Use the elimination method to
solve the simultaneous equations three 𝑎 plus two 𝑏 equals 14, four 𝑎 plus
two 𝑏 equals 16.
So we have a pair of
simultaneous equations or a system of linear equations in two variables 𝑎 and
𝑏. And we’re told that we must use
the elimination method in order to solve this system of equations. We’ll label our two equations
as equation 1 and equation 2 for ease of referencing them. And looking at the two
equations, we notice, first of all, that they have exactly the same coefficient
of 𝑏. They both have positive two
𝑏. Now your first thought may be
that we can, therefore, eliminate the 𝑏-variable by adding the two equations
together. But let’s see what that looks
like.
On the left-hand side, three 𝑎
plus four 𝑎 gives seven 𝑎. We then have positive two 𝑏
plus another positive two 𝑏, which gives positive four 𝑏. And on the right-hand side, we
have 14 plus 16, which is equal to 30. So we have the equation seven
𝑎 plus four 𝑏 equals 30. This equation still involves
both variables, so we haven’t achieved our aim of eliminating one, which means
that adding the two equations together wasn’t the correct step to take.
Instead, let’s try subtracting
one equation from the other. And as the coefficient of the
other valuable, 𝑎, is greater in equation 2 than it is in equation 1, I’m going
to try subtracting equation 1 from equation 2. On the left-hand side, four 𝑎
minus three 𝑎 gives 𝑎. We then have two 𝑏 minus two
𝑏. So that cancels out to
zero. And on the right-hand side, 16
minus 14 is two. So we have 𝑎 equals two. We’ve eliminated the
𝑏-variable. And in fact, we found the
solution for 𝑎 at the same time. The correct way to eliminate
one variable then was to subtract one equation from the other. And the reason for this is that
the coefficients of the variable we were trying to eliminate, that is the 𝑏’s,
are identical in both equations and they have the same sign.
There is a helpful acronym that
we can use to help us remember this, SSS. It stands for if we have the
same signs, then we subtract. We must remember that it is the
signs of the variable we are looking to eliminate that is important. So it’s the 𝑏’s that we were
interested in here. As the signs of the 𝑏’s were
the same, we eliminated them by subtracting one equation from the other.
Now that we found the value of
𝑎, we need to find the value of 𝑏, which we can do by substituting our value
of 𝑎 into either of the two equations. Let’s choose equation one. We have three multiplied by two
plus two 𝑏 is equal to 14. That’s six plus two 𝑏 equals
14. And subtracting six from each
side gives two 𝑏 is equal to eight. We then solve for 𝑏 by
dividing each side of the equation by two, giving 𝑏 equals four.
So we have our solution to the
simultaneous equations: 𝑎 is equal to two and 𝑏 is equal to four. But we should check our answer,
which we can do by substituting the pair of values we found into the other
equation. That’s equation 2. Substituting 𝑎 equals two and
𝑏 equals four into the left-hand side of equation 2 gives four times two plus
two times four. That’s eight plus eight, which
is equal to 16, the value on the right-hand side of equation 2. So this confirms that our
solution is correct.
We need to remember then that
helpful acronym SSS, which stands for if the signs of the variable we want to
eliminate are the same, then we subtract. Of course, the reverse is also
true. If the signs of the variable we
want to eliminate are different, then we add. Our solution to this set of
simultaneous equations which we’ve checked is 𝑎 equals two and 𝑏 equals
four.
Now, in the two examples we’ve seen
so far, we’ve been able to eliminate one of the variables straightaway by adding or
subtracting the original two equations. Sometimes, though, there may be an
extra step needed before we can do this, which we’ll see in our next example.
Using elimination, solve
simultaneous equations five 𝑥 minus four 𝑦 equals 21, four 𝑥 plus 12𝑦 equals
32.
So we’re asked to solve this
system of equations using the elimination method, which means we’re looking to
eliminate either the 𝑥- or 𝑦-variable by adding or subtracting our two
equations. However, if we were to try this
as the equations currently are, we’d find that in both cases we still have 𝑥-
and 𝑦-variables in the equation we are left with. If we were to add, we’d have
the equation nine 𝑥 plus eight 𝑦 equals 53. And if we were to subtract,
we’d have the equation 𝑥 minus 16𝑦 equals negative 11. So this hasn’t actually
helped.
So why hasn’t it worked? Well, in order to use the
method of elimination, we’re looking for the coefficients of one of the
variables to be the same in both equations, or at least to have the same
magnitude such as positive and negative three. But in this problem, this isn’t
the case. We have a coefficient of five
for 𝑥 in the first equation and four in the second. And we have a coefficient of
negative four for 𝑦 in the first equation and 12 in the second. So simply adding or subtracting
these equations as they currently are doesn’t eliminate either variable.
We’re told, though, that we
need to use the method of elimination. So what should we do? Well, what we’re going to do is
manipulate these equations slightly so that we do have the same coefficient, or
at least the same magnitude of coefficient for one of the variables. To achieve this, we use the
equality property of multiplication, which says that if we multiply both sides
of an equation by the same value, then the equality is still true. We’re looking for a value that
we can multiply one equation by which will then give the same coefficient, or
same-size coefficient, for one of the variables in both equations.
Looking at the 𝑦-variables in
our two equations, we see that four is a factor of 12. So if we were to multiply
equation 1 by three, then we would have negative 12𝑦. And so the magnitude of the
coefficients of 𝑦 would be the same in both equations. Let’s try that then. Let’s multiply equation 1 by
three. So we multiply everything in
equation 1 by three, giving 15𝑥 minus 12𝑦 is equal to 63. The coefficients of 𝑦 in our
two equations are now the same size but with different signs, which means we can
eliminate the 𝑦- variables by adding our two equations together.
When we do, we have 15𝑥 plus
four 𝑥, which gives 19𝑥; negative 12𝑦 plus 12𝑦, which gives zero; and on the
right-hand side 63 plus 32, which is 95. So we’ve eliminated the
𝑦-variable from our equations, giving a single equation in 𝑥, which we can
solve by dividing both sides by 19. Doing so gives the solution for
𝑥. 𝑥 is equal to five.
We can then find the value of
𝑦 by substituting 𝑥 equals five into any of the three equations, either of the
original two or the equation we created when we multiplied equation 1 by
three. I’m going to use equation 2 as
all of the coefficients are positive in this equation. Doing so gives four times five,
which is 20, plus 12𝑦 equals 32. We can then subtract 20 from
each side and divide by 12 to give 𝑦 equals one. So we have our solution: 𝑥
equals five and 𝑦 equals one. But of course, we should check
it. I’m going to choose to check by
substituting the values into equation 1. That’s five 𝑥 minus four 𝑦
equals 21. Substituting 𝑥 equals five and
𝑦 equals one gives five times five minus four times one. That’s 25 minus four, which is
indeed equal to 21. So this confirms that our
solution is correct.
The key stage in this question,
then, was to notice that we couldn’t eliminate either variable by adding or
subtracting the two equations in their original form. Instead, we had to first multiply
one equation by a constant to create an equivalent equation in which the coefficient
of 𝑦 was the same magnitude as it was in the second equation. Only then could we use the method
of elimination to solve these simultaneous equations.
Now it may not always be possible
to just multiply one equation by a constant. Instead, we may need to multiply
both equations by different constants. Let’s have a look at an example of
this.
Using elimination, solve the
simultaneous equations four 𝑥 plus six 𝑦 equals 40 and three 𝑥 plus seven 𝑦
equals 40.
In the two equations we’ve been
given, the coefficients of 𝑥 are different and the coefficients of 𝑦 are also
different, which means we can’t eliminate one variable by just adding or
subtracting the equations together. We also notice that neither of
the coefficients of 𝑥 are factors of the other and neither of the coefficients
of 𝑦 are factors of each other. We want to create equations in
which the coefficients of either variable are the same or at least the same but
with different signs. So how are we going to do
this?
Well, we’re going to have to
multiply both equations by some constant. I’m going to choose to multiply
equation 1 by three and equation 2 by four because this will create 12𝑥 in each
equation. We could also have chosen to
multiply equation 1 by seven and equation 2 by six as this would create 42𝑦 in
each equation. It doesn’t matter which
variable we choose to eliminate.
Now that we have our two new
equations, we have 12𝑥 in each. And as the signs are the same,
we can eliminate the 𝑥-variable by subtracting. I’m actually going to subtract
the top equation from the bottom one because the coefficient of 𝑦 is greater in
the second equation. We have 12𝑥 minus 12𝑥, which
cancels out; 28𝑦 minus 18𝑦, which gives 10𝑦; and 160 minus 120, which is
40. So we’ve eliminated the
𝑥-variable from our equation. We can then solve for 𝑦 by
dividing both sides of this equation by 10, giving 𝑦 equals four.
To solve for 𝑥, we substitute
this value of 𝑦 into any of our four equations. I’m going to choose equation
1. Doing so gives a
straightforward linear equation for 𝑥 which we can solve by subtracting 24 and
dividing by four to give 𝑥 equals four. So we have our solution. Both 𝑥 and 𝑦 are equal to
four. As always, we should check our
solution by substituting our values into any of the other equations. I’ve used equation 2. And it confirms that our
solution is correct.
The key step in this question
then was to multiply both equations by some number to create the same
coefficient of one of the variables. We could then use the method of
elimination to eliminate this variable and solve our simultaneous equations.
Let’s review some of the key points
from this lesson. Our aim in this method is to
eliminate one variable by adding or subtracting the two equations. If the coefficients of the variable
we wish to eliminate have different signs, we add the two equations together. And if the signs are the same, then
we subtract the equations, which we can remember using the acronym SSS. We also saw that in some cases we
may need to multiply one or both equations by a constant before we can add or
subtract to eliminate one variable.
Once we’ve eliminated one variable
and found the value of the other, we need to substitute this value back into one of
our equations in order to find the value of the variable we eliminated. And we should always check our
answer by substituting both values into whichever equation we didn’t use when
working out the second value.
