# Bunch ofÂ examples

## Video transcript

Let's do a couple of example
problems where we use information to set up equations
to solve our problem, and then we'll actually
solve the equation. So let's do problem number 1. Write a mathematical equation
that describes the relationship between the
variables in the table. So you have your days
and your profit. And if we look at them, on day
1 profit of 20, on day 2 profit of 40, day
3 profit of 60. It looks like no matter what
happens the profit is always going to be equal to-- let's
see, it's 20, 40, 60, 80. It looks like every day
it increases by 20. This is plus 20, to go from 20
to 40, then it's plus 20, plus 20, plus 20. So every day it increases
by 20. So maybe the formula here is
that our profit is equal to the day times 20, is equal
to 20 times the day. Does that fit the data? Let's see, you could try it with
all of them, but when the day is 5, 20 times 5 is 100. 5 times 20 is 100. Yeah, it looks like it actually
fits the data. And if we wanted to write it a
little bit more algebraically, we could say that profit is p
and day is d, and we would write our equation as profit is
equal to 20 times d or 20d. So that's part a. I wrote a mathematical equation
that describes the relationship. And then part b, they say what
is the profit on day 10? Well lucky for us we've written
our equation, so they're essentially just saying
what happens to p, profit, what happens to profit
when d or day is equal to 10? So profit is going to be equal
to 20 times the day we're on, day 10. 20 times 10, which
is equal to 200. That's it. Problem 3. Write a mathematical equation
for the following situations and solve. All right, let's do part a. 7 times a number is 35. What is the number? So whenever someone says a
number, I'll just use the variable x. I could use any variable,
I could say y or b or z or anything. But I'll go with x because
that's what I'm most used to and what's most typically
done unless you've already used your x. So 7 times a number, I'll say
7 times x is equal to 35. So that's our equation. We've written our mathematical
equation. And now it's, what
is the number? So to solve for x you could
it one of two ways. You could divide both sides
of the equation by 7. So if you divide that side by 7
and that side by 7-- And the whole reason I'm doing that
is because I want this coefficient to become 1. And anything I do to one side
of the equation I have to do to the other side. So 7 divided by 7 is 1, so you
just have 1x or you just have x is equal to 35 divided by 7. x is equal to 5. The other way you could have
thought about this, you could have said that this is
multiplying both sides of the equation times 1/7. So 1/7 times 7x is equal
to 35 times 1/7. These are equivalent
ways of doing it. Either way these would
have canceled out. You would have gotten
1x is equal to 5. Part b. One number is 25 more than
2 times another number. All right, so let's say
that x is equal to a number, the one number. And let's say that y is equal
to the other number. So they're introducing
more variables here. So one number, let's say x,
is-- is, that's like equal --it's equal to 25 more than
2 times another number. Well, this is the
other number. So 2 times our other number. That's what I got out of this
first line right there. One number is 25 more than
2 times another number. 2 times another number plus 25
is going to be equal to my first number. All right, now what is the
second [UNINTELLIGIBLE]. If each number is multiplied by
5, their sum would be 350. So that means 5 times this plus
5 times that is going to be equal to 350. So 5 times x plus 5 times y is
going to be equal to 350. And they're asking what
are the numbers? Well, here we can do something
called substitution. We already see that x is equal
to this whole thing over here. It's just like saying x is equal
to 5 or x is equal to 9 or whatever. x is equal to something. So every time we see an
x, we can replace it with this something. So we see an x right here, so
we can replace it with this something, and then we'll have
an equation that only has a y in it and then we'll be
able to solve for y. So this can be written as 5
times, instead of an x we have all this business, 25 plus 2y. I substituted for x, because
they told us x is equal to this thing. 25 plus 2y plus 5y
is equal to 350. Use the distributive property. 5 times 25 is 125 plus 5 times
2y is 10y, plus all of this, 5y is equal to 350. 10y plus 5y is 15y. That's 15y plus 125
is equal to 350. Now we can subtract 125 from
both sides, so minus 125-- Sorry, minus 125. I'm doing that to get
rid of the 125 from the left-hand side. So our equation simplifies to--
I'll continue it up here. We get 15y is equal
to 350 minus 125. That is equal to 225. And you may or may not already
know that 15 times 15 is 225, so we already know what y is. Let's divide both sides
by 15 just to make it clear what I'm doing. You divide both sides by 15 you
get y is equal to 225 over 15, which is equal to 15. Now we're not completely
done yet. We know what y is, but we still
have to solve for x. So x is equal to 25
plus 2 times y. y is equal to 15. 2 times 15, that
is equal to 30. 25 plus 30 is 55. Let me do part c. Let me clear this
out of the way. Clear this out of
the way as well. Part c. The sum of two consecutive
integers is 35. What are the numbers? Well, let's say that x is equal
to the first integer. Then what's the second
integer going to be? Well the second integer's
just going to be one more than that. They're consecutive, they're
one after another. So x plus 1 is equal to
the second integer. And they're saying the sum of
both of these numbers is 35. So x plus the number that comes
right after x, x plus 1, is equal to 35. Add these together, you get
2x plus 1 is equal to 35. Subtract 1 from both sides of
the equation, you get 2x is equal to 34-- just subtracted
1 from both sides. Divide both sides by 2 and
you get x is equal to 17. So the first number is 17, and
the second number is just one more than that, it's 18. And notice, two consecutive
integers, 17 then 18, add them together, you get 35. Part d. Peter is 3 times as old
as he was 6 years ago. How old is Peter? So Peter-- I'll say p for
Peter's age --is equal to 3 times as old as he
was 6 years ago. How old was he 6 years ago? Well, if he's p years old
now, he was p minus 6 years old 3 years ago. So let me write this as p minus
6-- that's as old as he was 6 years ago. That's what p minus 6 is. This 3 times-- that's the
3 times as old as he was 6 years ago. And, of course, this is how
old he is right now. So we just have to
solve for p. So we get p is equal to,
distribute the 3, 3 times p is 3p minus 3 times 6 is 18. Now we can subtract 3p from both
sides of the equation. We get minus 2p, right? p
minus 3p is minus, or negative, 2p is equal
to negative 18. And divide both sides by
negative 2 and you get p is equal to 9. Just multiply both sides
by negative 2. And you can verify it. Right now he's 9. 6 years ago he was 3, and he
is 3 times as old as that. Right, 6 years ago he was 3,
and at 9 years old he is 3 times as old as he
was 6 years ago. So it definitely works out. That's the cool thing about
algebra, you can always check your work. You always know when you
got the right answer. Problem 6. The price of an MP3 player
decreased by 20% from last year to this year. This year the price of
the player is $120. What was the price last year? So let me say, p is equal
to the price last year. And they're telling us that
the price of an MP3 player decreased by 20% from last
year to this year. So let's take 20%
away from this. So price minus 20% of the
price-- this is decreasing by 20% --is equal to the price
this year, which they're telling us is $120. Well what's p minus 0.2
or minus 20% of p? We decreased by 20%. Well you're going to
be left with 80%. So we could say 80% of the price
last year is equal to the price this year. Like we did before, we divide
both sides by the coefficient on our variable. Divide both sides by 0.8, or
0.80, depending how you want to view it. 0.80. These cancel out. So we have to divide
0.8 into $120. A little exercise in
dividing decimals. Let me put some zeroes there. We could multiply the 0.8
and the 120 by 10. That's essentially moving
the decimals over to the right by one. So this is the same as dividing
8 into 1,200. So this decimal isn't there
anymore, it's over here. This decimal is now over here. 8 goes into 12 one time,
1 times 8 is 8. 12 minus 8 is 4, bring
down the zero. 8 goes into 40 five times. 5 times 8 is 40, 40 minus 40
is 0, bring down another 0. 8 goes into 0 zero times, 0
times 8-- you get the idea. 0 times 8 is 0. We have no remainder. So our answer is $150. So it was $150 last year,
it decreased by 20%. 20% of $150's $30, so it
decreased by $30, and it definitely got us to $120.