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# Simple equations: examples solving a variety of forms

Video transcript

Let's do some examples
of solving equations. And you're going to see the
equations in this video, they're not going to require a
lot of steps, but I really want to focus on what's actually
happening on the equation, and why it makes sense
to do the steps that we're doing. So this first equation here,
we have x-- I'll rewrite it --x plus 11 is equal to 7. So that's telling us-- And I
have this little dot here, let me eliminate, let me
clear that dot. Don't want you to think I have
some type of a decimal there. There you go. So back to the problem. So what is this saying? This is saying, some number
plus 11 is equal to 7. And you may already be able
to do this in your head. You know, what do I have to
add to 11 to get to 7? So it's going to have
to be some type of a negative number. But we're going to learn how
to do this systematically. So what we want to do is, when
we solve an equation, it should be of the form,
when we're all done, x is equal to something. And that's going to
be our answer. We'll know what number satisfies
this equation. Now, to get there, what we want
to do is to just have this x on the left-hand
side of the equation. We don't want this 11 there. We want to get rid of this 11
somehow, on the left-hand side of this equation. Now to do that, we can't just
get rid of it only on the left-hand side. You might want to just say, hey,
let me subtract 11 from the left-hand side
of the equation. But this is an equation. Anything you do to one side of
the equation, you have to do to the other side
of the equation. If x plus 11 is equal to
7, x plus 11 minus 11 won't equal 7 anymore. You'll have to also subtract
11 from the 7. Now they're equal. Remember, I want to really
emphasize this. I want you to really
think about what this equality means. Equation, it comes from the
notion of equality. This is equal to this. These two things are equivalent
quantities. Now, in order for this equal to
hold true, anything I do to this, if I'm subtracting 11 from
this quantity, in order for the equality to hold true,
I also have to subtract 11 from this quantity right here.
x plus 7 and 7 are equal. So they're still going to be
equal after I subtract 11. Now let's focus on the
left-hand side. Negative 11 plus x plus 11,
negative 11 plus 11, they're going to cancel out, right? 11 minus 11 is 0. So you're going to be with left
with just an x is equal to 7 minus 11. Now what's 7 minus 11? You can draw a number line
out, if you like. If we draw a number line right
here-- 0, 1, 2, 3, 4, 5, 6, 7. 7 is there. And if you subtract 11 from
that, you go back 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. 8, 9, 10, 11. That takes you to negative 4. 7 minus 11 is negative 4. That's our first equation. Now we have 7 times some
number-- this is the new problem --7 times some number
x is equal to 21. This is going to be a different
x than this one right here. So how, once again, do we get
it in the form of x is equal to something? Remember, we've got x is
equal to something. How do we get into that form? Let me rewrite the problem. 7 times something
is equal to 21. Well, I could think
of it two ways. If I divide 7 times
x by 7, what am I going to be left with? If I divide that by 7, what am
I going to be left with? 7 times something divided by
7, that's just going to be that something. That's just going to
end up with an x. But if I do it on the left-hand
side of the equation, I have
to do it on the right-hand side of the equation. Remember, 7x is equal to 21. It's the same thing as 21. In order for this to be equal to
that when I divide by 7 on the left, I also have to divide
by 7 on the right. And when I do that, 7 times x
divided by 7-- that's just going to be x --and then 21
divided by 7 is equal to 3. So the solution to this equation
is x is equal to 3. And you can verify it. Take 3, plug it into
that equation. 7 times 3 is indeed 21. Same thing over here. Take x is equal to negative 4. Negative 4 plus 11 is the
same thing as 11 minus 4, is indeed 7. Let's do this one here. This one looks a little bit
more difficult, but you're going to see, it's
not too bad. So we have 5x over 12
is equal to 2/3. Now, how do you isolate
the x over here? What can we multiply or divide
by to isolate it? Well, what if we were to
multiply both sides of this equation by 12 over 5? What's going to happen? So if I multiply the left-hand
side by 12 over 5-- And remember, anything I do to one
side of an equation, I have to do to the other side
of an equation. So if I multiply the left side
by 12 over 5, I have to multiply the right side by 12
over 5 in order for that equality to hold true. But why would I pick
12 over 5? Because I wanted to pick a
number that cancels perfectly with the 5 over 12. Right? I have a 5 over 12 here. 12 over 5 times 5 over 12, the
12's cancel out, the 5's cancel out, and you're
left with just an x. So you get x is equal to
2/3 times 12 over 5. You can divide the 12
by 3 and get a 4. You can divide the 3
by 3 and get a 1. So you get 2 times 4 is
equal to 8 over 1 times 5, which is 5. So x is equal to 8/5. Now you might say, hey, Sal,
over here you divided by 7, here you're multiplying by the
inverse of the coefficient. When do I know what to do? And the answer is, you
can do either. I could have done this exact
same problem that way. I started with 7x
is equal to 21. Instead of dividing by 7, I
could have said, let me multiply by the inverse of 7. So I could have multiplied by
the inverse of 7 on both sides of the equation. This is equivalent
to dividing by 7. You can verify it. When you multiply by 1/7, that's
the exact same thing as dividing by 7. So if you work this out,
these would cancel out. This is 1/7 times 7/1. The 7's cancel out. You get x is equal to 21 over
7, which is equal to 3. Likewise, over here,
I have 5x over 12. Let me write it this way. I could write it as, 5 over
12x is equal to 2/3. This is the exact
same problem. I'm just separating the
x out a little bit. 5/12 times x is 5x over 12. Now here, I could just say, let
me divide both sides by the coefficient 5 over 12. So I could have just divided
by 5 over 12. But what does it mean to
divide by 5 over 12? That's the same thing as
multiplying by 12 over 5. Dividing by a fraction is the
same thing as multiplying by its inverse. So this and this, we're doing
the exact same thing. Whether you divide by the
coefficient, like we did over here, or whether you multiply
by the inverse of the coefficient, you're doing the
exact same operation. Now let's do some of these other
problems. And this is really just to show you that the
variable isn't always x. We can solve for q. We can solve for z. We can solve for s. We can solve for anything. So let's solve these
down here. So we have-- and I'll rewrite
it --you have q minus 13 is equal to negative 13. So once again, we want
to isolate the q. We want it to be, q is
equal to something. Then we will have solved
our problem. So how can we isolate the q? How can we get rid of this 13? What if we add 13 to both
sides of this equation? And I'm adding 13, because if
I add 13 to negative 13, or minus 13 right here,
I'm going to get 0. So the left-hand side of this
equation is going to become q, and then what is the right-hand
side, right? That and that will cancel out. And the right-hand
side, negative 13 plus 13, is just 0. So our solution is,
q is equal to 0. And we can check it. 0 minus 13 is indeed
negative 13. So it works. Actually I forgot to check
this one over here. Let's check it. The fun thing about
algebra, you can always check your answer. We got x is equal to
8/5 over here. Let's check our answer. 5 over 12 times x times 8
over 5 is equal to what? 5's cancel out. You can divide the 8
by 4 and get a 2. You could divide the 12
by 4 and get a 3. It equals 2/3. So we can verify. That's the neat thing. Once you get your answer, you
can always make sure, if you have enough time, that you
got the right answer. Let's do these last two. We have z plus 1.1 is
equal to 3.0001. Once again, we want to get
rid of this 1.1 on the left-hand side. The best way to do that that I
can think of is subtracting 1.1 from both sides, or adding
negative 1.1, however you want to do it. If I do it on the left, I've
got to do it on the right. Otherwise the equality won't
still hold true. If this is equal to this, in
order for the two sides to be equal, anything I do
to the left, I have to do to the right. So let's see. If I subtract 1.1 from z plus
1.1, those two cancel out, and I'm left with just a z is equal
to 3.0001 minus 1.1. So here we're going
to do a little bit of subtracting decimals. That's probably the hardest
part of this problem. So 3.0001 minus 1.1. You want to align
the decimals. Minus 1.1. Just like that. And we can even add some 0's
there, just to make it clear. And let's do our subtraction. 1 minus 0 is 1. 0 minus 0 is 0. 0 minus 0 is 0. Uh oh, we don't have-- We
can't do 0 minus 1. We can regroup, or
we can borrow. So this will become a 10,
which is really a 1. And we're going to take
a 1 from here, so this becomes a 2, right? We borrowed from the 3, we
regrouped the numbers. We took the whole 1 and put it
in the tenths place, so it becomes 10/10. So 10 minus 1 is 9. We have our decimal space. And then 2 minus 1 is 1. So our answer is z is
equal to 1.9001. And we can verify that. What's z plus 1.1? Let's add it up. You get a 1, 00, 9.1 is 10, 1
plus 1 plus 1 is 3, we have our decimal sign, and
we got 3.0001. That's a good review of
subtracting decimals. All right. And we have this last problem. 21s is equal to 3. Like we said, we could do
this two ways, which are essentially the same way. 21 times some number
is equal to 3. Well, what happens if
we multiply both sides by 1 over 21? By the inverse of
the coefficient. The coefficient is just the
number that sits out in front of the variable, that's
multiplying the variable. Well, 1 over 21 times 21 is just
1, so we get s is equal to 3 over 21. And 3 over 21 is the same
thing as 1 over 7. And we can verify that. What is 21 times 1 over 7? It's equal to 21 over 7. 21 divided by 7. Which is 3. And just to really hit the point
home, I could have done this problem-- Instead of
multiplying by 1 over 21-- let's see, 21s is equal to 3
--I could've, instead of multiplying by 1 over 21, a
completely equivalent thing to do would have been to just
divide both sides by 21. So these would cancel out. You'd get s is equal to 3
over 21, which is 1/7. Anyway, hopefully you found
those examples useful.