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# Ex 3: Distributive property to simplify

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. Ć’ 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.