 This video is going to talk about formulas and how to solve a formula for a variable and then how to work with them in word problems. So the first thing that you want to do is always clear your fractions and then you want to clear the parentheses and then you can apply those addition and multiplication properties. So you look at it and I want to solve for R. I have no fractions but I do have parentheses so I need to multiply everything by that P. I need to distribute the P. So on this side I have A and then P times one is P and then P times RT is just PRT. Well I want to solve for this R so I've got to make sure that I get this R on one side by itself and then if I have to take things away I can't. So I only have one R is on one side so I'm ready to start doing the addition property and the addition property tells me that I want to take this P and take it to the other side. I've got to peel that layer of my onion so we subtract P from both sides. Now remember A and P are different. It's like apples and pears. So I have to say A minus P is equal to PRT. Well now I'm looking at this problem and I want to get to that R again but I've got this P and this T that are being multiplied by it and how do you get rid of something that's being multiplied by your variable? You divide it. So I need to divide this side by PT and I have to divide the whole other side by PT and when you're working with fractions you can just divide the whole side. So the P's and the T's become factors of one and we're just left with R equal to A minus P on the top divided by P over T and I'm good with that answer. You don't have to take it any further than that. Let's try this formula. I'm trying to solve for C. I got a fraction so let's clear the fraction. You wouldn't necessarily have to clear the fraction because I just have a constant in that thing but we were going to follow the rule so we're going to clear the fraction. So if I clear the fraction I need to multiply everything by that denominator of five so I have to multiply. Let's look at it this way. I'm going to distribute the five to everything. I typically don't write it this way because I'll forget and it gets messy. I like to cancel things out but we can rewrite this as 5 times F and then equal to 5 times 9 over 5 times C plus 5 times 32. If I write the 5 times everything then I won't forget. Okay this is still 5F equal to this 5 is a denominator and remember this is like over 1 so it's a numerator and so I have 9C left plus 5 times 32 and 5 times 32 is 160. Now again what are we solving for? We're solving for C so I need to get rid of this 160. It's being added so to peel that layer off I'm going to subtract 160 and I can't combine it literally with my 5F so I just have to rewrite 5F and then subtract that 160 and then is equal to my 9C and I'm trying to go for C so I'm going to divide by 9 because it's being multiplied by 9 and my final answer is 5F minus 160 over 9 is equal to my C. So this example is the same kind of problem except that we have a y equal 1 half x minus 8. It doesn't really look like a formula but this is the kind of equation that we're going to be solving in the next chapter or two. So let's see what we could do for it. We want to solve it for x excess. So here's what we're trying to solve for and to do that then I need to peel this 8 off so I'm going to add 8 to both sides and remember I can't combine my y and my 8 so I have y plus 8 on the left hand side and that's equal to 1 half x. Now we had a fraction here and last time we cleared the fraction but this time since I just had one term with a fraction in it and I figured I would have to clear it. I could just work with it and see what happens. So I had this 1 half x it's being multiplied by x so that means I need to multiply both sides by 2 over 1 and that means both whole sides have to be multiplied by 2. So I have 2y plus 16 when I distribute my 2 and that's going to be equal to the 2 over 2 becomes a 1 and I just have x. So let's look at how we can translate and then use a formula. The area of a triangle with a base of 18 inches and a height of 7 inches we're trying to find that area but we need to know what the formula is and the formula for area of the triangle is 1 half the base times the height. So let's see what we have. We want to find the area so that's a and the base b is 18 inches so I have 1 half times 18 inches times the height h which is 7 inches so 7 inches and I need to figure out what that is. Well we are talking about multiplication here and so that means inch times inch when I'm thinking about my units and remember when you multiply the same thing over and over again we can use an exponent and the exponent tells me that I have two factors of inches so inches squared is going to be our units. So now that we know what the units are going to be we can really just rewrite this as 1 half times 18 times 7 so a is going to be equal to two things at a time 1 half times 18 would be 9 times 7 and 9 times 7 is going to be 63 and from over here we know that that's inches squared is the area. So when you're multiplying your units will have an exponent on it. This one's a little trickier but let's read it. Find the length of a rectangle that's what we're looking for and we're going to call that l of a rectangle if the perimeter that should say the perimeter here we're going to call that p is 64 inches and the width we're going to call that w is 20 inches. Now why did I put all that in there because the formula is perimeter distance around is equal to two times a length plus two times the width. Now let's think about this for a minute we're going to have our length and inches plus our width and inches and if I add inches and inches they're the same unit. I'm not multiplying this time so I don't have factors I'm just this is almost like saying apples plus apples I'm just totaling my inches so if my unit is going to be inches when you add and subtract you keep the unit if you multiply you're going to have an exponent on your unit. So we have p is 64 is equal to two times the length well length is what we're going to be solving for plus two times the width which happens to be 20. So there's our equation and we have one variable so that's nice and if we keep going then 64 expanding this out would be 2l plus 40 and now we know that we need to subtract the 40 to peel off the constant layer and we have 24 is equal to 2l and then if we divide that by two we're going to find out that the length is equal to 24 which means that in a sentence the length is 24 inches.