## Video transcript

Let's do a few more examples
of solving equations. And I think you're going to
see that these equations require a few more steps
than the ones we did in the last video. But the fun thing about these is
that there's more than one way to do it. But as long as you do legitimate
steps, as long as anything you do to the left-hand
side, you also do to the right-hand side, you should
move in the correct direction, or you shouldn't
get the wrong answer. So let's do a couple of these. So the first one says-- I'll
rewrite it-- 1.3 times x minus 0.7 times x is equal to 12. Well, here the first thing that
my instinct is to do, is to merge these two terms.
Because I have 1.3 of something minus 0.7 of
that same something. This is the same variable. If I have 1.3 apples minus 0.7
apples, well, why don't I subtract 0.7 from 1.3? And I will get 1.3 minus 0.7
x's, or apples, or whatever you want to call them. So is equal to 12. You could imagine that I did
the reverse distributive property out here. I factored out an x. But the way my head thinks about
it is, I have 1.3 of something minus 0.7 of
something, that's going to be equal to 1.3 minus 0.7 of those
somethings, that x. And of course 1.3 minus 0.7 is
0.6 times x of my somethings is equal to 12. And now, this looks just like
one of the problems we did in the last video. We have a coefficient times x is
equal to some other number. Well, let's divide both sides
of this equation by that coefficient. Divide both sides by 0.6. So the left-hand side will
just become an x. X is equal to-- and what
is 12 divided by 0.6? 0.6 goes into 12-- let's add
some decimal points here-- that's the same thing
as 6 going into 120. 6 goes into 12. 2 times 2, times 6 is 12. Subtract, you get a 0. 6 goes into 0 0 times. So it's going to go 20 times. 12 divided by 0.6 is 20. And we can verify. Let's verify this. 1.3, let's substitute it back. 1.3 times 20, minus
0.7, times 20. Let's verify that that
is equal to 12. So I'll take the calculator out,
just so you don't have to trust my math. So we have 1.3 times
20 is equal to 26. So this piece right
here is 26. And then 0.7 times 20. I don't need a calculator
for that. That is 14. 26 minus 14 is 12. So it checks out. We got the right answer
for this equation. x is equal to 20. Let's do this one right here. 5x minus 3x plus 2
is equal to 1. This looks very complicated. And whenever something looks
daunting, just do steps that look like they're simplifying
the equation. And over time, as long as you
do legitimate steps, you should be able to make
some progress. So the first thing I want to
do, is I want to distribute this negative 1 over here. So this is the same thing
as 5x minus 3x, minus 2. Right? I just did the distributive
property on the 3x and the 2. This is a negative 1
times 3x plus 2. So it's negative 1 times 3x,
plus negative 1 times 2. Or negative 3x minus 2. And that is going to
be equal to 1. Now, I have 5 of something
minus 3 of that same something. So that's going to be equal
to 2 of that something. 5x minus 3x is 2x. 5 minus 3 is 2. And then I have the minus
2 is equal to 1. And now, I like to get into the
form where I have 2x, or I have something times x is
equal to something. So I want to get rid of
this negative 2 on the left-hand side. The best way I know how
to do that is to add 2 to both sides. So add a 2 on the
left-hand side. If I do it to the left-hand
side, I've got to do it to the right-hand side. Plus 2 on the right-hand side. These two guys will cancel out,
and you're going to get 2x is equal to 1 plus
2, is equal to 3. And now you can divide both
sides by 2, and you get x is equal to 3/2. And I'll leave it for you to
verify that this is indeed the correct answer. Let me draw a little line here
so that our work doesn't get messy, although that might have
made it even messier. So here we have to
solve for s. And look, we have a fraction
and 2 s terms. How do we do that? Well, just do it the same way. We have 1 times s minus-- you
can view this as 3/8 times s is equal to 5/6. You could view this as 1 times
s, minus 3/8 times s is equal to 5/6. You could factor out
an s, if you like. Maybe I'll do it this way. I'll factor it onto the
left-hand side. This is the same thing
as s times 1, minus 3/8 is equal to 5/6. And 1 minus 3/8, what is that? That 1, I can rewrite as 8/8. That's 1. So this is the same thing as 8/8
minus 3/8 is 5/8, times s. You could switch the order
of multiplication. 5/8 times s is equal to 5/6. And you might be able to
go straight from that. If I have 1 of something minus
3/8 of that something, I have 8/8 of that something minus
3/8 of that something, I'm going to have 5/8 of
that something. And now, to solve for s, I can
multiply both sides by the inverse of this coefficient. So I multiply 8 over
5 times 5/8s. If I do it to the left-hand
side, I have to do it to the right-hand side. 8 over 5. I multiplied by 8 over 5 so
that those cancel out and those cancel out. And you are left with s is equal
to-- right, this is just a 1-- is equal to-- well,
the 5's we can divide. Divide the numerator and
the denominator by 5. Divide the numerator by 2 and
the denominator by 2. You're left with-- sorry, divide
the denominator by 2, you get 6 divided by 2 is 3. You're left with 4/3. s is equal to 4/3. Let's do one more of these. So here I have 5 times q minus
7 over 12 is equal to 2/3. So let me write this. And I could rewrite this as just
5 over 12 times q minus 7 is equal to 2/3. And what I want to do with this
video, is to show you that I can do it two
different ways. But as long as I do legitimate
operations, I should get the same answer. So the first way I'm going
to do it, is I'm going to multiply both sides of
this equation by the inverse of 5/12. So I'm going to multiply both
sides by 12 over 5. Because I wanted to get
rid of this 5/12 on the left-hand side. It makes everything look
a little bit messy. And I multiply it by 12 over 5,
because these are going to cancel out. The 5 and the 5 cancel out, the
12 and the 12 cancel out. So the left-hand side of my
equation becomes q minus 7 is equal to the right-hand
side, 2/3 times 5/12. If you divide the 12
by 3, you get a 4. You divide the 3 by
3, you get a 1. So 2 times 4 is 8 over 5. And now we can add 7 to both
sides of this equation. So let's add-- I want to do that
in a different color-- add 7 to both sides
of this equation. These two 7's cancel out. That was the whole point
of adding the 7. And you are left with q is
equal to 8/5 plus 7. Or we could write 8/5 plus
7 can be written as 35/5. And so this is going to be
equal to 8-- well, the denominator is 5. 8 plus 35 is 43. So my answer, going this way,
is q is equal to 43/5. And I said I would
do it two ways. Let's do it another way. So let me write the
same problem down. So I have 5/12-- actually, let
me just do it a completely different way. Let me write it the
way they wrote it. 5 times q minus 7, over
12 is equal to 2/3. Let me just get rid of the 12
first. Let me multiply both sides of this equation by 12. I just don't like that 12
sitting there, so I'm going to multiply both sides by 12. So these are going to cancel
out, and you're going to be left with 5 times q minus 7
is equal to 2/3 times 12. That's the same thing
as 24 over 3. So this is, let me write this. 2 over 3 times 12 over 1 is
equal to-- if you divide that by 3, you get a 4, divide
that by 3, you get a 1-- is equal to 8. So you get 5 times q minus
7 is equal to 8. And then instead of dividing
both sides by 5, which would get us pretty close to what we
were doing over here, let me distribute this 5, I just want
to show you, you can do it multiple legitimate ways. So 5 times q is 5q. 5 times negative 7 is minus, or
negative 35, is equal to 8. 5q minus 35 is equal to 8. Now, if I want to get rid of
that minus 35, or that negative 35, the best
way to do it is to add 35 to both sides. I did that so that these cancel
out, and I'm left with 5q is equal to 8 plus
35, which is 43. Now I can multiply both sides of
this equation by 1/5, which is the same thing as dividing
both sides by 5. And these cancel out. You get q is equal
to 43 over 5. So there's a bunch of ways you
can do these problems. But as long as you do legitimate
steps, you will get the right answer. And I'll leave it to you to
verify that this truly is the right answer for q. This is the q that will
satisfy this equation. Let's do one word
problem here. Jade is stranded downtown with
only $10 to get home. Taxis cost $0.75 per mile, but
there's an additional $2.35 hire charge. Write a formula and use it to
calculate how many miles she can travel with her money. All right. So the total cost of a cab ride
is going to be equal to just the initial hire charge,
which is $2.35, plus the $0.75 per mile, times the
number of miles. We're letting m is equal to
the miles she travels. Miles traveled. So this is the equation. We know that she only
has $10 to get home. So her cost has to be $10. So we have to say, the
cost has to be $10. So 10 is equal to
2.35 plus 0.75m. So how do we solve for
m, or the number of miles Jade can travel? Well, we can get rid of the 2.35
on this right-hand side by subtracting that
amount from both sides of this equation. So let's do that. So let's subtract minus
2.35 from both sides. These will cancel out. That was the point. The left-hand side-- what
is 10 minus 2.35? Now, these will cancel out. Now what is 10 minus 2.35? 10 minus 2 is 8. 10 minus 2.3 is 7.7. So it's going to be 7.65. If you want to believe
me, let's do it. 10 minus 2.35. 7.65. And that is going to
be equal to 0.75m. Let me write that in
that same color. It's nice to see where different
things came from. 0.75m. I have, like, five shades
of this purple here. so this is that, that is that,
and then these two guys canceled out. Now to solve for m, I can just
divide both sides by 0.75. So if I divide that side by
0.75, I have to do it to the left-hand side as well. 0.75. That cancels out, so on the
right-hand side, I'm left with just an m. And on the left-hand side-- I'll
have to get my calculator out for this one-- I have 7.65
divided by 0.75, which is equal to 10.2. m is 10.2, so Jade can
travel 10.2 miles.