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All right.
We're going to solve some formulas for a specified
variable here.
And what I mean by that is we're going to have a formula
like the one on the left, ax equal to b.
And we're going to solve in this case for x.
So we don't have any numbers.
We just have letters.
And the way I want to think about these is treat the
letters-- like the a and b that we're trying to move away
from the x--
treat them just like numbers and do the same thing you
would do with numbers.
So if I had instead 5x equals 10, I'd want to get rid of
this 5 that's with the x here.
It's being multiplied on to x, so I'll divide it off.
Divides out and I get x is equal to 2.
I'm going to do the exact same thing with the letters that I
don't want next to the variable. a is being
multiplied to x.
So I divided it off.
We usually do the opposite operation of what's attaching
it to the variable.
That gives me x is equal to b/a.
And that would be our final answer because we can't
simplify that at all.
All right.
Let's try out another one.
Let's say we have something being added to
the x in this case.
So x plus a is equal to b.
And again, treat the a and b like numbers.
Use the same process.
So if instead I had something like x plus 5 is equal to 8.
and I want to isolate the x.
So in this case, again, we're solving for x.
Over here with numbers, if I want to get rid of the plus 5,
I subtract it.
So it combines to 0.
And I have to balance it out by doing it on the other side.
So that gives me x is equal to 8 minus 5 or 3.
I'm going to do the same thing with the a and the b here.
x plus a.
I'm solving for x.
So I want x to be by itself.
I need to get rid of the a.
So I subtract it off because it's being added.
Then it combines to 0.
I have to subtract a on both sides.
And we can't combine the b and the a here, so we're just
going to write it out.
x equals b minus a.
And it doesn't really look like an answer, but it is.
That is the solution for x if we're given x plus a equals b.
And let's try out another one.
Let's see.
I'm running out of space here.
Let's move this up.
OK.
Now let's try out something with a bit more
happening and not x.
This time, we'll use some different variables.
And we're going to use the same idea.
Let's solve in this case for g.
All right.
So if we're solving for g, we're going to treat every
other letter or variable like just a number and do whatever
we would do to a number.
So the first thing I want to do is isolate the entire term
that contains g--
because I'm looking for g--
on one side of the equation.
That means I want to get rid of this term here, plus 3v.
If it's being added to the term I'm looking for, I
subtract it off, get it to combine to 0.
So it cancels out.
I'm going to subtract 3v over here.
And with the s minus 3v, there's
nothing I can do to that.
So I just write s minus 3v equals-- bring down the term--
3g.
And again, we're looking for g.
So now we have a 3 being multiplied to it.
I don't want the 3 to be there.
I need to get rid of it, which means I'm going
to divide it off.
So it cancels or reduces down to 1.
And I have to divide the entire left side by 3 also.
And that gives me my final answer, s minus 3v over 3 is
equal to g.
We could also write this in a different format.
So I'll just say or--
and either answer is OK.
Actually, let's go this way.
If I give the 3 denominator to each term separately, or in
other words, up here, I divide each of these terms by 3
separately, I would have s over 3 minus 3v
over 3 equals g.
And then the 3/3 here, reduce down to 1.
So I have another format for the final answer, s over 3
minus v equals g.
All right.
So both of those are equivalent to the same thing.
And let's try one more that has parentheses in it.
So we're solving the usual process-- or we're following
the usual process for solving a linear
equation, exact same steps.
And we're just pretending that the letters that we don't
want, the letters we're not looking for, are numbers.
Treat them like numbers.
So let's say we have g is equal to 1/9 pi m squared x.
And let's say we're solving for x.
Well, this one has fractions.
So it's really nice to get rid of fractions.
So first step, let's multiply both sides of the equation by
this denominator of 9 to make the fractions go away.
So if I do 9 times, 9 times.
And that's what we usually do in linear equations, right?
We try to get rid of fractions first.
So if I multiply both sides by 9, this reduces to 1.
And remember, you can think of 9 as 9 over 1.
So you're really multiplying two fractions.
Reduce, and we get a 1 there.
So now we have 9g is equal to-- and what's left over is
pi m squared x.
And again, we're looking for x.
And now we look and see, well, there's nothing being added or
subtracted here.
But we do have pi m squared being multiplied onto x.
If it's being multiplied, and I don't want it there, I need
to divided it off.
So remember to do opposite operation.
If I divide it off, that reduces to 1.
That reduces to 1.
And I have to balance it by dividing the exact same thing
on the other side.
And then we have our final answer, 9g over pi m squared
is equal to x.