>> [Whispers] Math is fun and you can do it.
This is Part 2 of solving literal equations which means you've got a formula with more
than one variable and you're solving for one of the variables.
So this is a tax formula, T equals 3/10 I minus 12,000, we're going to solve for I.
So first you notice there's only I in one place and it happens to be in the parentheses.
So there are two ways you can go about doing this,
one is to try to simplify the right hand side so it's only got what's in parentheses first
and you can do that by multiplying by the reciprocal
of what's being multiplied outside the parentheses.
So we have 3/10 times something.
So we can multiply both sides by 10/3 so watch what happens if I do 10/3 on each side,
we would have 10/3 times T and 10/3 times 3/10 times I minus 12,000 and when we do
that then the 10/3 times 3/10 cancels so that what I have
on the left side is 10/3 T or 10 T over 3.
It doesn't matter which way you write it.
I'm going to write as 10/3 T equals and since these cancel I just have I minus 12,000.
You don't need the parentheses any more.
Now I'm trying to solve for I, so we have I minus something
so what we could do is add 12,000 to both sides to isolate I, so this gives you 10/3 T
and if I add 12,000 to both sides and solve for I, this is what I get, right.
Lastly I would suggest you write the I on the left hand side,
so we have I equals 10/3 T plus 12,000, so I solved for I.
Now you could write this a little bit differently, you could write that as 10 T
over 3, you can get a common denominator and if you got a common denominator
by multiplying the right side by...the 12,000 by 3
over 3 you might see it written even a little bit differently,
it's possible that you might write this as 10 T plus...now if I multiply the top and bottom by 3
over 3 that would give you 36,000 over 3, so then the whole thing would be over 3
so that's another way of getting the same answer.
All right so these are both correct, you know let's look at the original again and let's do it
by doing the distributive property first.
Okay, so I'm going to take the same problem and we're going to go down
and try it a different way so same problem but I'm going to do the distributive property first
and see what happens, because there's usually more than one way to solve for a variable.
So if I do the distributive property I have 3/10 I, and again you could write that as 3 I over 10
or 3/10 I, I think it's easier to just see it
as the coefficient minus...now what do I have here, 3/10 times 12,000.
Okay. So we have 3/10 times 10,000, so I can either leave that with a common denominator
and write 36,000 over 10 or I can cancel so again, more options.
For the moment I'm just going to leave it with a common denominator.
Even though of course 36,000 over 10 is the same thing as 3,600,
so those of you who wrote 3,600, that's correct too.
So then I'm going to...I've done the distributive property, now what I'm going
to do is multiply both sides by the least common denominator so we've covered...that's a way
to get rid of the fractions in an equation.
So since the least common denominator is 10 and,
by the way that's why I left it this way you'll see in a minute, I can multiply both sides
of the equation by 10, okay I'm going to move this over just a little bit.
All right, so if I multiply both sides by 10 remember you have
to multiply it to every term, also by this.
So I have 10 T equals...now I want to do 10 times 3/10 I, that will give me 3 I
because the tens will cancel which is the whole idea of multiplying
by the least common denominator and 10 times the last term over here,
the tens will cancel and I have 36,000.
By the way if I had just written this as 3,600 and then multiplied
by 10 I would get back to 36,000.
Right now we're going solve for I so I have to add 36,000 to both sides, right,
because I want to isolate the variable I, so I have 10 T, add 36,000 to both sides
and now I've got to divide by 3 to isolate the I so the last thing is to divide by 3.
So over on the left hand side that's what I equals, I'm just going to write it
as I equals 10 T plus 36,000 all over 3.
All right so that's a different approach to solving the same problem
and let's just go back up, 10 T plus 36,000 over 3...ah...remember
that was another way I could have written it before.
Remember I had 10/3 T plus 12,000 and by putting it
over a common denominator I also got 10 T plus 36,000 over 3.
Here is a geometrical formula, the V stands for volume R is the radius of a circle,
the volume equals 4/3 pi R cubed, actually the sphere, not a circle.
All right so how we going to solve for pi...well basically there's only pi in one place
and it's easier if you see the coefficient of pi written before it, so you could write V equals,
now we have 4/3 and we also have an R cubed so what you could do is write 4/3 R cubed
or write 4 R cubed over 3 times pi, it's easier to see the coefficient that way.
And if you multiply by the reciprocal of this fraction in front of pi everything will cancel
and you'll have it solved for pi.
So we're going to multiply both sides by the reciprocal which is 3 over 4 R cubed and times V
on this side and times 4 R cubed, 3 times pi on this side.
So on the right side everything cancels except the pi right, so we've solved for pi
and on the left side we have 3 V over 4 R cubed.
So pi equals 3 V over 4 R cubed.
Okay, you could have done this in two steps by on the second line here you could get rid
of the fraction first by multiplying both sides by 3 and then
at the second step divide both sides by 4 R cubed, but if you multiply
by the reciprocal it's done in one step.
Okay here's another one, V equals X minus Y over M and we're going to solve for M,
all right since this contains a fraction you can multiply both sides
by the least common denominator which is M, all right so let's do that.
If I multiply both sides by M, so on the left side I would have M times V or I'm going
to write that as V M, it doesn't matter if you write M times V or V times M
and on the right hand side the M's cancel so I have X minus Y. Now I want to solve for M,
there's only M in one place, right so I just need to divide by the coefficient of M,
so if I divide both sides by V, I'm done, M equals X minus Y
over V. Now the original problem V could have written...been written as V
over 1 equals X minus Y over M and you could have used proportions
if you remember how proportions work, you multiply what's on one diagonal, the V times M,
this only works when you have an equation when they're both fractions on each side.
I could write V times M is equal 1 times X minus Y
and of course notice you get exactly the same thing as I had
by multiplying by the least common denominator.
So you'll still get M equals X minus Y over V. All right so here...that's a lot
of examples solving literal equations, practice, practice, practice, have fun.
[Whispers] I hope you learned something, remember math is fun.