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Learn about Applications of Division Equations Video
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 Learn about Applications of Division Equations Video
TenMarks teaches you how to apply division equations to solve real life problems.
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Learn about Applications of Division Equations Division equations. So the first problem it states that a builder wants to install 10 solar cells on the roof of the new house. Each solar cell is 8 feet wide. How wide must the roof be to fit all of the cells? So, in this problem, we can use a division equation to solve. Now, in order to this, we’re going to follow these steps. So our first step is in setting up our division equation is to write to equation. So we’re going to let R equal or represent the width of the roof. What we’re looking for. How wide the roof must be, so R is going to equal our unknown, the width of the roof. So R division equation is going to look like such. We’re going to take the width of the roof and we’re going to divide our width of the roof by the width of one solar cell. And that would give us the number of solar cells. So now, let’s go ahead and plug in the information we know. So the width of the roof, we don’t know. So we’re going to put that with the variable R our unknown. So there’ll be R divided by the width of one solar cell is 8 feet wide and the number of solar cells is 10. So now, we have our equation set up. So our second step is we need to isolate the unknown. So we are isolating our variable, our unknown, okay. Now to isolate the unknown, we’re going to use inverse operation. Now, because we are dividing the inverse operation of the division is multiplication. Inverse operation is the opposite, so the opposite of division is multiplication. So we need to multiply both sides of this equation to isolate the unknown. And if we multiply both sides of the equation, the equation will still be true. So I'm going to rewrite my equation and I'm going to go ahead and I'm going to multiply each side by eight. So I'm going to take R divided by eight times eight and then I'm going to take ten times eight. When I do that, my eights cancel out and I'm left with R equals and 10 x 8 is 80. Now, so the width of the roof is 80, before we move on though, let’s check our answer. We always want to check your answer to make sure you are correct. So to check your answer, all we’re going to do is we’re going to substitute 80 for R. So we’re going to take our original equation, R divided by eight equals ten and I'm going to plug in 80 for R. So 80 divided by 8 equals 10. And then I'm going to go ahead and solve, so 80 divided by eight is ten equals ten. So since both sides of the equation are the same is the answer is correct. So therefore, the roof must be at least 80 feet wide. So the roof must be 80 feet wide. Let’s move on to our next problem. So in this problem, we need to solve M divided by five equals two thirds. So M divided by equals two thirds is a division problem. So this is a division problem. Now to solve the equation, we need to isolate the unknown. We have to isolate the M, okay. So again, we need to isolate our unknown which is M in this case. All right, so to isolate the unknown, we’re going to use inverse operation. So inverse operation remembers is the opposite, so if we’re doing division, if we’re dividing the opposite of division is multiplication. So if we multiplied both sides by the same number, this equation will be true. So I'm going to take my problem, M divided by five equals two thirds and I'm going to multiply both sides by five. So I'm going to multiply M divided by five, I'm going to multiply that by five and I'm going to multiply two thirds by five. So when I multiply M divided by five by five, the fives cancel out and I'm left with M. and then, if I multiply two thirds by five, I'm left with two times five is ten thirds, right which an improper fraction. So if I change it to proper fraction, I get three goes into ten, three times which is nine with one leftover. So the next number is three and one thirds. So M equals three and one third. Things to remember and to keep in mind is that a quantity on the left side of the equation. So a quantity on left side of the equal sign is ba