Video Transcript
In today’s lesson, what we’re gonna
look at is applications on systems of linear equations. But what does this mean? Well, it means that we’re gonna
take a look at real-life situations, and then we’re gonna form systems of linear
equations from these. And then we’re gonna solve them to
find our variables, which may be 𝑥 and 𝑦 or any other letters you choose.
Now, what we’re gonna have a look
at is problems that have different amounts of variables, so it might have one
variable, it might have two variables. And we could look at different
methods to solve them. And we could use skills such as
simultaneous equation solving. So to help lead us into the
problems, what we’re gonna have a look at is an example that only has one
variable. So let’s take a look at this
now.
A man’s age is four times his son’s
age. In five years, the sum of their
ages will be 105, how old are they now?
So in this problem, we can see that
we could actually use two variables or one variable to solve it. Well, in fact, this is a good
example because we see it and think there’re two things there. But we can use only one variable if
we’d like to solve the problem. And this in fact makes the answer
simpler to come to. So what we done first of all is
we’ve selected a variable, and the variable is gonna be 𝑥. And what this is is the son’s
age. Could’ve been any letter, I just
decided to use 𝑥.
Okay, well, if the son’s age is 𝑥,
then the man’s age must be four 𝑥 because the man’s age is four times his son’s
age. And what these variables will
represent are the man and his son’s age now. But we’ve also been told that we’re
looking at their ages in five years’ time. So we could also then look at our
expressions for ages in five years’ time. First of all, the son’s age. Well, the son’s age is just gonna
be 𝑥 plus five. That’s because it’s in five years’
time, so we add on five. And therefore, the man’s age is
gonna be four 𝑥 plus five cause, again, we’re just adding five to his age now.
So now, to be able to form an
equation, we can use the other bit of information we’re given, and that is that the
sum of their ages in five years’ time is going to be 105. So therefore, we can form our
equation, which is 𝑥 plus five plus four 𝑥 plus five equals 105. So now, on the left-hand side of
the equation, what we can do is collect like terms. We’ve got 𝑥 add four 𝑥, which is
five 𝑥. And then we’ve got five add five,
which is 10. So therefore, we’ve got five 𝑥
plus 10 equals 105.
So now what we need to do is
subtract 10 from each side of the equation. So when we do that, we get five 𝑥
equals 95, and then we divide both sides of the equation by five, which will give us
𝑥 is equal to 19. So therefore, we can say that the
son is 19 years old. So now, what I want to do is find
out how old the man is now. Well, the man’s age is represented
by four 𝑥. So therefore, four 𝑥 is equal to
four multiplied by 19, as 𝑥 equals 19. So therefore, this would give us
76. So what we can say is that the son
and the man’s ages now are 19 years old and 76 years old, respectively.
So, great! In this problem, we had a look at
something that had one single variable. We could’ve set out with two. But actually, sometimes you don’t
need to. And this was a good way to
introduce us to how we would turn a real-life situation into an equation. Now, what we’re gonna look at is
problems with systems of equations, so they’re gonna contain more than one
variable.
So when we do have a problem that
has two variables, what we’ll do is set up a couple of equations, these are called
simultaneous equations. And we can use various methods to
solve these, which we’re going to show you. And as we say, if we got two
variables, they could be 𝑥, 𝑦; 𝑎, 𝑏; 𝑙, 𝑚. It doesn’t matter as long as we’re
consistent throughout the problem. So now, what we’re gonna have a
look at is our first example involving two variables.
I am thinking of two numbers. Use the clues to determine what the
numbers are. When you divide one by the other,
the quotient is eight. The sum of the two numbers is
81.
So in this problem, what we have
are two variables because we’re told that we need to find two numbers. So therefore, what we’re gonna call
these variables is 𝑥 and 𝑦. So now what we can do is set up a
couple equations using our two clues. We’ll use the first clue. We’re told that if we divide one by
the other, the quotient is eight. So therefore, we can set up our
first equation which is 𝑥 divided by 𝑦 is equal to eight. And that’s because the quotient is
the result of a division. And I’m gonna label this equation
one cause this helps us as we go through our method.
Now, we can take a look at the
second clue and set up an equation using this because we’re told that the sum of the
two numbers is 81. So therefore, we have 𝑥 plus 𝑦
equals 81. And we’ve called this equation
two. So now, what we’ve got is a pair of
simultaneous equations. So what we’re gonna use to solve
these simultaneous equations is a method called substitution. And to do this, we need to
rearrange one of our equations so that 𝑥 or 𝑦 is the subject. Well, in fact, you could use either
of the equations, but what we’re gonna use is equation one. And we could rearrange this by
multiplying each side of the equation by eight. And when we do this, we’ve got 𝑥
as a subject to the equation cause 𝑥 is equal to eight 𝑦. And we’re gonna call this one
equation three.
So as we said, we’re gonna be using
the substitution method. So what we can do now as we’ve got
𝑥 as the subject is substitute equation three into equation two. As I said, we could’ve made 𝑥 or
𝑦 the subject and we could’ve rearranged either equation one or equation two. It’s just that we chose to do
equation one. So now that we’ve substituted in
equation three into equation two, we get eight 𝑦 plus 𝑦 equals 81, which is gonna
give us nine 𝑦 is equal to 81. So then, we divide each side of the
equation by nine and we get 𝑦 is equal to nine. So great! What we’ve done is we’ve found our
first variable or, in fact, our first one of our two numbers.
So now to enable us to find our
other variable 𝑥, what we’re gonna do is we substitute 𝑦 equals nine into one of
our other equations. Well, in fact, the best one to
substitute it into is equation three because we already got 𝑥 as the subject. And when we do this, what we get is
𝑥 is equal to eight multiplied by nine. So this’s gonna give us 𝑥 is equal
to 72. So therefore, we can say that our
values are gonna be 72 and nine.
And what we can do is we can
double-check this by substituting them back into our original equations that we got
from our clues. So if we look at our first clue or
our first equation, we’re gonna have 72 divided by nine is equal to eight. Well, yes, this is correct because
72 divided by nine is eight. Well then for our second equation,
we got 𝑥 plus 𝑦 equals 81. Well, we’ve got 72 plus nine equals
81. So again, this is correct. So therefore, yes, 72 and nine are
our two numbers.
So okay, great! We’ve solved a problem that has two
variables, and we’ve used simultaneous equations. However, these were fairly simple
because our coefficients of our 𝑥- and 𝑦-variables have been one. Let’s now take a look at some more
complex problems where this isn’t the case.
In a test with 20 questions, 𝑥
marks are awarded for each correct answer and 𝑦 marks are deducted for each
incorrect answer. Benjamin answered 12 questions
correctly and eight questions incorrectly, and he scored 44 points. Emma answered 14 questions
correctly and six questions incorrectly, and she scored 58 points. How many points were deducted for
each incorrect answer?
So in this question, we can see
that we’ve got two variables. So what we’re gonna want to do is
use the information we’ve been given to set up two equations and solve them
simultaneously. And then what we’re looking to find
is what our 𝑦 is because we want to find how many points were deducted for each
incorrect answer. Well, the first bit of information
we can use to form the first equation is the fact that Benjamin answered 12
questions correctly and eight questions incorrectly and he scored 44 points.
So first of all, we’re gonna begin
with 12𝑥, and that’s because we know that there are 𝑥 marks for each correct
answer and Benjamin answered 12 questions correctly. And then this is gonna be minus
eight 𝑦. And we get that because we’re told
that 𝑦 marks are deducted for each incorrect answer. So therefore, we get eight 𝑦
because there’re eight questions incorrectly answered by Benjamin. However, the reason it’s minus
eight 𝑦 is it’s because 𝑦 marks are deducted. So we’re taking them away. So now, we’ve got an expression
12𝑥 minus eight 𝑦. And then this is equal to 44 cause
we’re told that Benjamin scored 44 points overall. Okay, great! So I’ve labeled this equation one
cause this will help us when we’re identifying what to do in the next steps.
Now, let’s look at equation
two. Well for equation two, we’re gonna
have 14𝑥 minus six 𝑦 equals 58. And that’s cause we’re told that
Emma answered 14 questions correctly, so that’s 14𝑥, and six questions incorrectly,
so we subtract six 𝑦. And then the total score that she
got was 58 points. Okay, great! So now what’s our next step? So what we want to do now is solve
our equations simultaneously. And the method we’re gonna use is
elimination. However, to use elimination, what
we need to do is we need to eliminate either 𝑥 or 𝑦. And to do that, we have to have the
same coefficient of either 𝑥 or 𝑦. It doesn’t matter if the signs
aren’t the same, it just has to be the same value in front of our 𝑥 or 𝑦.
So this isn’t the case with our
equations. So therefore, what makes this a
slightly more complex problem is the fact that we now need to find a number that we
need to multiply each of our equations by to enable us to have the same coefficient
of either 𝑥 or 𝑦. Now, what we can do is we can
multiply equation one by three and equation two by four because what this is going
to do is give us a coefficient in both cases of 24 for our 𝑦. Well, in fact, it’ll be negative
24. And that’s because eight multiplied
by three is 24 and six multiplied by four is also 24.
So we’re gonna start with equation
one. And if we multiply this by three,
what we’re gonna get is 36𝑥 minus 24𝑦 equals 132. I’m gonna call this equation
three. And then if we take a look at
equation two, if we multiply this by four, we’re gonna get 56𝑥 minus 24𝑦 equals
232. And we’ve called this equation
four. And it’s worth noting at this
point, be careful of a common mistake. And that is where people forget to
multiply each of the terms. And most likely, it’s the term on
the right-hand side of our equation, so the numerical value they forget to
multiply.
Well, now, if we take a look at the
coefficients we’ve got, we’ve got 24𝑦 in both of our equations. And in fact, we also have them both
as negative, so they have the same signs. So to eliminate the values, what
we’re gonna do is same-sign subtract. And that’s cause if we have
negative 24 minus negative 24 is the same as negative 24 add 24, which is just
zero.
However, it’s worth mentioning here
a common mistake cause people often just look at the central signs of the equations
and then deal with those. So if in this case, they’re both
the same. However, if you’re trying to
eliminate the 𝑥-terms, then it would in fact be the signs associated with the
coefficient of our 𝑥-terms which will be the ones that you would look at. So now, to enable us to eliminate
one of our variables, so in this case 𝑦, what we’re gonna do is subtract equation
three from equation four. And when we do that, what we get is
20𝑥 is equal to 100. And that’s because 56𝑥 minus 36𝑥
is 20𝑥. And then, as we said, we already
eliminate the 𝑦-terms. And then we’ve got 232 minus 132,
which is 100. So then we divide through by 20,
what we get is 𝑥 is equal to five.
So great, we found out what 𝑥
is. However, this is not what the
question is looking for cause it wants us to find out how many points were deducted
for each incorrect answer. And this is the variable 𝑦. So then, to find 𝑦, what we do is
substitute 𝑥 equals five into equation one. We could choose any equation. I just happen to have chosen
equation one here. So when we do that, what we’re
gonna get is 12 multiplied by five minus eight 𝑦 equals 44, which is gonna give us
60 minus eight 𝑦 equals 44. So then, what we can do is
rearrange to solve to find 𝑦. And so what we do is subtract 44
from each side of the equation and add eight 𝑦. And we do this so that we have a
positive 𝑦-term.
So then what we get is 16 is equal
to eight 𝑦. So then if we divide through by
eight, we’re gonna get two is equal to 𝑦. So therefore, as we found out that
𝑦 is equal to two, we can answer the question cause we can say that two points were
deducted for each incorrect answer. And it would be possible to check
our answer by substituting our 𝑥- and 𝑦-values into any one of our four
equations.
So great! We’ve seen a variety of problems,
with those with one variable to those with two variables and those, now, with a more
complex simultaneous equations to solve. For our final example, what we’re
gonna take a look at is a problem that combines skills. So we’re gonna take a look at
something that’s got percentages and our systems of equations.
Jacob has 20,000 dollars to
invest. His intent is to earn 11 percent
interest on his investment. He can invest part of his money at
eight percent interest and part at 12 percent interest. How much does Jacob need to invest
in each option to get a total 11 percent return on his 20,000 dollars?
So the way that we’re gonna solve
this problem is have a look at the values that are involved. So the first thing we’re going to
do is work out how much Jacob is gonna have if he gets his intended 11 percent
interest. Well, to work out what would happen
if we increase 20,000 dollars by 11 percent, what we’re gonna do is multiply 20,000
by 1.11. And we get the 1.11 multiplier
because if we think about 100 percent, well, 100 percent is equal to one. Well, if we add on 11 percent, then
it’s gonna be 111 percent. And 111 percent is equal to
1.11. And we get this because percent
means by the 100 or out of 100. So what we do is we divide our
percentage by 100 to get the decimal. So that would give us our 1.11,
which is gonna give us 22,200. So this is gonna be our final
amount after investment that jacob is looking for.
So now what we’re gonna do is work
out what the two investments are. But the two investments are gonna
be 𝑥 and 𝑦 cause that’s the variables we’ve decided to use, with 𝑥 representing
the investment at eight percent interest and 𝑦 representing the investment at 12
percent interest. So then, to set up our first
equation, we’re gonna get 1.08𝑥 plus 1.12𝑦 equals 22,200. And that’s because after the 11
percent interest, we get the 22,200. And then we get the 1.08𝑥 because
we have an increase of eight percent. It’s like if it’s a multiplier of
1.08. And then we’ve multiplied that by
the investment amount, which is 𝑥. Then, we get the 1.12 again,
similarly, because we’re increasing by 12 percent. And we’ve called this equation one
cause this is gonna help us when we identify what the next steps are.
So then, for equation two, what
we’re gonna get is 𝑥 plus 𝑦 equals 20,000. And that’s because we know there
are two investments and the total that jacob has to invest is 20,000 dollars. So therefore, 𝑥 plus 𝑦 is gonna
be equal to 20,000. So now, the best way to solve these
simultaneous questions is to use the substitution method. And that’s because we’ve got a
simple second equation, which we can rearrange easily to make 𝑥 or 𝑦 the
subject. So what we’ve chosen to do is make
𝑥 the subject. So to do this, we’re gonna subtract
𝑦 from each side of the equation. So when we do that, we get 𝑥
equals 20,000 minus 𝑦, which we’re gonna call equation three.
So then, in order to use our
substitution method, we’re gonna substitute our new equation for 𝑥 into equation
one, so substitute equation three into equation one. So once we’ve done this, the next
step is to distribute across our parentheses. So what this is gonna give us is
21,600 minus 1.08𝑦 plus 1.12𝑦 equals 22,200. So now, we can simplify this by
subtracting 21,600 from each side of the equation, which is gonna give us 0.04𝑦
equals 600. And then, if you divide each side
of the equation by 0.04, you get 𝑦 is equal to 15,000.
So now, we found 𝑦, what we can do
is substitute this into equation three to find 𝑥. So when we do this, we get 𝑥 is
equal to 20,000 minus 15,000, which is gonna give us an 𝑥-value of 5,000. So therefore, we can say that the
investment that’s gonna be needed is 5,000 dollars at eight percent and 15,000
dollars at 12 percent.
So great, we looked at a range of
examples. So now let’s have a quick look at
the final summary of key points. So, in order to solve story or
real-life application problems, we need to use the following method. First of all, choose variables to
represent unknowns in the question. Doesn’t matter what the variables
are, 𝑥, 𝑦, 𝑎, 𝑏, etcetera. This doesn’t matter. Two, use the variables, translate
words to mathematical equations. Then, solve the equations. And then finally, express your
answer in terms of unknowns from the problem, not the variables we introduced. And that’s really important. So not “𝑥 is this, 𝑦 is
this.” Use the unknowns from the problem
itself.
