# 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.