Current time:0:00Total duration:9:28
0 energy points
The power of the distributive property allows us to solve these examples equations (and a word problem). Created by Sal Khan and CK-12 Foundation.
Video transcript
Like the last video, I want to start with two warm up problems. And then we'll do an actual word. And you're going to see these are going to be a little bit more involved than the equations in the last video. But we're still going to be doing the exact same operations, or what we could consider legitimate operations, to get our answer. So here we have 3 times x minus 1 is equal to 2 times x plus 3. So let's see what we can do here. The first thing I like to do-- and there's no definite right way to do it-- there's several ways you could do these problems-- but I like to distribute out the 3 and the 2. So 3 times x minus 1, that's the same thing as 3x minus 3-- I just distributed the 3-- is equal to-- distribute out the 2. 2 times x, plus 2 times 3, which is 6. Now what I like to do is get all of my constant terms on the same side of the equation and all my variable terms on the same side of the equation. So let's see if we can get rid of this 2x term on the right hand side. So let's subtract 2x. I'm going to do a slightly different notation this time because you might see it done this way. Or you might find it easier to visualize it this way. It doesn't matter. It's the same thing we did in the last video. But I want to subtract 2x from this side of the equation. But if I subtract 2x from the side of the equation, I also have to subtract 2x from that side of the equation. So then when we subtract 2x from both sides of the equation, what do we get? Here we get 3x minus 2x. That's just 1x, or x minus 3. 2x minus 2x is no x's, or 0. Then you just have the 6. So we get x minus 3 is equal to 6. That was by getting rid of the 2x from the right hand side; subtracting it from both sides of this equation. Now we have this negative 3 on the left hand side. I don't want it there. I just want an x there. So to get rid of that we can add 3 to both sides of this equation. You could imagine this is being adding this equation to the equation 3 is equal to 3. 3 is, obviously, equal to 3. Negative 2x is, obviously, equal to negative 2x. You could do it either way. But if you add 3 to both sides of this equation the left hand side of the equation becomes just an x, because these two guys cancel out. x equal, and 6 plus 3 is 9. And we are done. And we can even check our answer. 3 times 9 minus 1 is what? This is 3 times 8. This is 24. So that's what the left hand side equals. What does the right hand side equal? That is 2 times 9 plus 3. That's the 2 times 12, which is also equals 24. So it all works out. x is equals to 9. Next problem. z over 16 is equal to 2 times 3z plus 1, all of that over 9. So it's a hairy looking problem. Let's multiply both sides of this equation by 9. So if you multiply both sides of this equation by 9, what do we get? We get 9 over 16z is equal to-- this 9 and that 9 will cancel out-- 2 times 3z plus 1. Now let's distribute this 2. So we get 9 over 16z is equal to 2 times 3z is 6z plus 2. Now let's get all of z's on the same side of the equation. So let's subtract 6z from both sides of the equation. So let's subtract minus 6z there, and that, of course, equals minus 6z there. And what do we get? On the left hand side, we get 9/16 minus 6. What is 6 if I have 16 as a denominator? 6 is equal to what over 16? Let me think about it. 60 plus 36. It's equal to 96/16. So it's 9 minus 96 over 16z. This is just 6-- I just rewrote minus 6 here-- is equal to-- these two cancel out. That's why I subtracted 6z in the first place. So it's going to equal 2. So what does this equal right over here? Let me do it in orange. 9 minus 96. Well let's see. The difference between 9 and 96 is going to be 87. And, of course, we're subtracting the larger from the smaller. So it's going to be negative 87 over 16z is equal to 2. And we're almost there. We just have to multiply both sides of this equation by the inverse of this coefficient. So multiply both sides of this equation by negative 16/87. These cancel out. 87, 87, 16 and 16. The negative signs plus. You're just left with z is equal to 2 times this thing. So 2 times negative 16 is negative 32/87. And we are done. That's not a pretty looking answer, but that's what we got. And you can try it with multiple methods. Maybe you can multiply both of the equation by 16 first. Maybe you can distribute out the 2 9's first. All sorts of things you can do. And you could also verify that this is, indeed, the answer. Let's do a word problem now that we're warmed up. All right. It says Manoj and Tomar are arguing about how a number trick they heard goes. Tomar tells Andrew to think of a number. Let's say that x is the number that Andrew thinks of. Think of a number. Multiply it by 5. So this is what Tomar is saying. ƒ is telling Andrew to think of a number, multiply it by 5, so let me do that. So multiply it by 5, and subtract 3 from the result. We'll do that in a different color. We'll do it in blue. Subtract 3 from the result. So subtract 3 from the result. Then Manoj comes along. So here's Manoj. Manoj comes along and tells Andrew once again to think of a number. So once again, think of a number. We'll call that x. So they both are telling him to think of a number. Add five. So now he's saying add five to the number. So you add 5 to number. And then multiply the result by 3. So multiply the result by 3. Andrew says that which ever way he does the trick, he gets the same answer. What was Andrew's number? So regardless of whether he does 5 times the number, and then subtracts 3, or whether he adds 5 and then multiples the whole answer by 3, he gets the same number. So that must mean that these two things are equal. He gets the same answer. That means that these two are equals. So let's solve this equation. We get 5x minus 3 is equal to 3 times x plus 5. A good place for me-- I like to distribute out this 3, so that is equal to 3x plus 15. Always have to remember to multiply the 3 times all of the terms in parentheses. This is a 15. And, of course, 5x minus 3 is equal to that. Now, I think you know how I like to operate. I like to get all of my ex-coeffients on one side. So let's get them all on the left hand side. Which means, let's get rid of them on the right hand side. So let's subtract 3x from both sides. So minus 3x minus 3x and then what do we get? Our equation becomes 5x minus 3x is 2x. You still have a minus 3 is equal to-- these cancel out-- equal to 15. Equal to that yellow 15 right there. Now we want to get rid of this negative 3 on the left hand side. And the best way to do it is to add 3 to both sides of the equation. So add 3 to both sides of this equation. So what do you get? You now get 2x-- and these cancel out-- is equal to 18. And then you divide both sides of this by 2. And you get x is equal to 18 over 2. Or 9. So in either situation, Andrew was thinking of the number 9. Let's see if it works according to both Tomar and Manoj's method. If I were to take 5 times-- if this is 9-- you get 5 times 9 is 45 minus 3. So you get 42. That's when you do Tomar's method. When you do Manoj's method-- this is a 9, right there. 9 plus 5 is 14. So this thing becomes 14, and it becomes 3 times 14. 3 times 14 is 30 plus 12, which is 42. So Andrew was right. When he randomly picked the number 9, and regardless of which method he uses, he gets the same results. He gets 42.