Algebra 1 – Module 5, Topic 1 – Solving Literal Equations  
Hi guys. Welcome to Algebra 1. Today's lesson's going to focus on solving literal equations. Now, if you think back to pre-algebra, you had some practice solving equations there, you're going to use that knowledge to get you through this one. You ready? Let's go.
Okay. Before we get into literal equations, what they are, let's think back to linear equations. Back in pre-algebra you probably solved something like this. You learned the steps to get through this. You'd see, okay, solve for x. You've got 5x=20, so you think, all right, you know when you're solving equations the goal is to isolate the variable. You want to get the variable all alone. If you look here, you have five times your variable, five times x, so you have to do the opposite or the inverse operation. In this case, of multiplying. The opposite of multiplying by five would be to divide by five, so you'd go through, you would divide both sides of your equation by five. On the left, the fives would cancel out, so you'd be left with x= and then twenty divided by five is four. I know in pre-algebra you learn to solve one step equations is what we will call those because we can solve them in one step. Okay? All right.
Continuing back on memory late, let's look at this. You could solve this in one step or two depending on how you thought about it. I'm going to show you how to solve it in two because I know it probably alarmed you a little bit that it had a fraction, but we're not afraid of those. Let's just dive in and I'll show you how to get rid of it, okay? Here, I have (2/3)x=4. What I want to do is first of all, get rid of that fraction, I know that would probably make most of us happy. Let's move this a little bit. Look at the fraction separately, consider it's numerator, consider it's denominator. We want to get rid of the denominator first. (2/3), if you thought about that like a division problem, it's two divided by three, and then we're multiplying that times x. Since I'm dividing by three, if I'm thinking about this as two divided by three, what's the opposite of dividing by three? Multiplying by three. That's how you'll start so you can get rid of this fraction here.  
I'm going to multiply the left side of my equation by three, put my little multiplication symbol and a three, and you know whatever you do to one side you've got to do to the other. Now I'm going to multiply the right side by three. On the left side, if you remember, your three's will cancel out. All you're left with is 2x, so 2x=. On my right side I have four times three. I know four times three is twelve, there, so now we're at this point where we just 2x=12. We probably feel comfortable with something like that, it's similar to the last one. You look here, okay the x, two is multiplying x, so what's the opposite of multiplying by two, divide by two, right? You would divide both sides by two, your twos would cancel out, so on the left side you're just left with x= and then twelve divided by two is six. You'd see x=6.  
I know in pre-algebra you handled those kinds, linear equations. Now look at this. Here, our equation is A=bh and they want us to solve for h. When we have a situation like this, we're solving a literal equation. Now, how you can tell the difference between a literal equation and a linear equation like the ones we were just handling is for literal equations your answer is not going to be a value, per say. It's not going to be x=4 or x=2 or x=75, your answer is some other algebraic expression, something else basically with variables still. Okay? Let me show you what I mean, let me work you through this one. 
Here I'm solving for h, and I have a=b times h, because I know when I see terms right together in algebra, multiplication is holding them together. I need to isolate h here. H is what I'm solving for. Pretend like they're just numbers, you're going to use the same process. I have b times h, so what's the opposite of multiplying by b? Divide by b. What is b, who knows, we don't know what b is, but we treat it the same way we would treat a number like we're doing the other equations. We're going to divide the right side by b, we're going to divide the left side by b. On the right side those b's cancel out and you're just left with h. You have h, we have, equals, and on the left we have a divided by b. You're done, for the most part, you've solved this. 
I know a lot of times it makes us feel more comfortable if we see the variable we were solving for on the left and the answer on the right, so you might want to reverse this. I'll say our answer is (a/b)=h or h=(a/b). Either one of those answers are acceptable to write, but you might feel more comfortable seeing the variable you were solving for on the left and the expression on the right, but they're both fine. 
Okay? You just solved your first literal equation. You see what I mean, how our answer, this a divided by b, it's not a numerical value because we don't know the values of a and b, they're variables, that's the only thing that makes literal equations different than linear equations. You'll look at your answer, you may feel like you're still not really done yet, but as long as you solve for the variable, you're done. 
Let's look at example two. Here we're solving for r and we have v=(pi)r(squared)h. A couple of things about this problem might have alarmed you a bit. First of all, you probably noticed the pi. I'm sure you've learned that you can approximate pi at about 3.14. It's actually a very long number that never ends and never repeats itself and it goes on and on and on for infinity. Your calculator will usually take it out about ten or eleven decimal places but generally we round it at 3.14. You'll notice the further along you get in your math classes in algebra one and geometry and algebra two that the further you keep going that you'll see pi more often then you'll see 3.14. When you see it, just use it, don't feel like you have to use the number instead, 3.14, just go ahead and keep going with pi. 
Another thing about this one, you probably noticed we have an exponent here. That's going to add a little bit to our steps when we're solving this problem. Generally, for the most part, the same process we're doing every time, okay? Let's jump in. First let me move this over a bit so I have a little more space to work, okay? I notice I have v=(pi) times r(squared) times h. I have the product of three terms on my right side. I know I'm solving for r so my goal is going to be to isolate r to get it by itself. Right now it's r squared so I'll work on getting that alone first. If all of these are being multiplied together, then I have to ask myself what's the opposite of multiplying, right? Dividing. Since I want r squared to stay but I want everything else to leave from around it, I'm going to divide by what I want to get rid of. I'm going to divide by (pi)h, but that product. 
Whatever I do to one side of the equation you'll have to do to the other, so I'm going to divide the left side by (pi)h. I've got that taken care of and what happens is on the right side your pi's will cancel, your h's will cancel so you're left with r(squared) and on the left side you're left with v/(pi)h. What is v? We have no idea, we don't know that either. Like we said, that's what makes it literal, because you're working with just a lot of variables in these kinds of equations, okay? 
Now I'm at the point where I have v/(pi)h = r(squared). I know that I'm trying to solve for r, not r(squared). I have to ask myself, okay, what's the opposite of squaring a number, because I know I need to undo the squaring. It's square root, right. We want to take the square root of our squared and the square root of v/(pi)h because that's going to allow us to just get at r, what we're trying to solve for. We'll take the square root of our squared and the square root of that quotient over there. The square root of r(squared), I know that's r. V/(pi)h, that square root, I'm just going to write it just like that, because I don't have numerical values for v and h. The square root of v/(pi)h. 
Now, like we said, a lot of times people are more comfortable seeing the variable they were solving for on the left and then the answer on the right, so you could rearrange this, I'll get this out of the way. You could write this as r= square root, big square root, v/(pi)h. Either one of those are acceptable answers, just which ever one you feel more comfortable writing. Okay? 
Let's try another one. See if you're starting to feel more comfortable with these. 
We have v=(1/3)bh.  In this one they're asking us to solve for b. We've got a fraction here but we practiced a little bit with that earlier, so we're not alarmed, right? If we're trying to get at b, we're trying to solve for b it's going to make us feel a lot better if we can just get rid of the fraction, right? Consider the denominator first. This case my denominator is three. If instead you think about that fraction, like one divided by three, that will give you a hint of your inverse operation. If you're dividing by three, what's the opposite of multiplying by three, right. I'm going to multiply the right side by three and I'm going to multiply the left side by three. On the right side, that allows my threes to cancel and I am just left with 1bh. Because one is not going to change anything as a coefficient, I am just going to write bh. 
Then I have three times v, so I have 3v. I'm not done yet, because I'm trying to get b. Here I have 3v=b times h, so h is multiplying times the b, so what's the opposite of multiplying by h? Dividing by h, right?  Divide the right side by h, divide the left side by h, and then what happens on the right is your h's cancel out and you're just left with b and then on the left you have (3v)/h and you are all done, got through that. What did it take about two steps. All right. 
It's your turn, guys. Go ahead and press pause and take a few minutes and work your way through these problems. Notice on each one the variable that I'm asking you to solve for and when you're ready to check your answers against mine, press play. 
Okay, let's see how you did. Get the pointer tool back. 
On the first one, you're solving for b. I got b=(2a)/h. On the second one we were solving for w. I got w=v/lh, okay? Stay with me, I'll show you how I got those, I'll show you my work. Let's get the pen, okay, here we go. 
I was solving for b, let's actually move a few things out of the way and get some room to work. Let's get that out of the way, and let's move this up, there we go. I want to get b by itself, so I look at my problem and I see what's going on with b. I think I'll scoot that over just a bit more too. First thing I see, I spot that fraction. Let's get that out of the way first. If I think about that fraction as one divided by two, that gives me a hint to get rid of it, or sometimes you'll hear people say clear their fraction, I need to multiply both sides by two. On the right, my two's cancel out and I'm just going to write bh, because that one as a coefficient isn't changing anything, and then on the left I have 2 times a, that's 2a, and I'm still trying to solve for b. I have 2a=b times h. The opposite of multiplying h, divide by h. Divide the right side and the left side by h. On the right side, your h's will cancel out so you're left with b and on the right, (2a)/h. 
Then, like you noticed on my answers, I reversed it when I wrote the answer.  This one is acceptable also, okay? 
Let me show you how I got that next one. Let's move some things out of our way. Move that, this time they're connected, okay. I'm going to rewrite that so I won't get that too crowded. 
V=l times w times h. Okay. I'm solving for w. I look at what's going on here. I have v=l times w times h. I want to get w by itself. It's being multiplied by l and h. What's the opposite of multiplying by l and h? Divide. I'll divide the right side by lh and then the left side, because whatever I do to one side remember I've got to do to the other. On the right, my l's cancel, my h's cancel, and I just have w. On the left, v/(lh). On the answers, I reversed these, but remember I said this is fine too. Whichever way you want to write it, okay?
All right guys. I hope you're feeling confident about how to solve literal equations. I hope to see here back here soon for more Algebra 1. Bye.