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Applied Linear Equations: Collection Problem - Concept
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
In Algebra II and in the real world, sometimes we need to solve collection math problems by using linear equations. When solving word problems using linear equations, we first need to pull out the relevant information and put it into equation form. When solving linear equation collection problems, we are usually asked about numbers of certain types of coins. These types of problems are sometimes called coin word problems.
Linear equations come in all shapes and sizes, for this particular example we are going to be dealing with a money problem. At 15 coins worth $3 if all I have is dimes and quarters how many of each denomination do I have?
Okay so we basically have a collection of change and we're trying to figure out how we can all make this work numerically. Let's take a look, so we have 15 coins worth $3 and I extend them like diagrams I make them all the time. So basically we know we have dimes and we know we have quarters okay. Dimes are worth 10 cents or a 10th of a dollar, quarters are worth 25 cents or a quarter of a dollar. And we don't know how many of either these, we have so let's say we have a certain number of times let's call it x okay. We don't know how many quarters we have either but we do know that we have 15 coins total. So if we have five dimes we know we'd have 10 quarters we have 7 dimes we have to have 8, so you subtract the difference from 15 not the number of quarters we have. Together we add these up and we end up with $3.
The trick that people sometimes have problems with is understanding the difference between the amount of money a coin is worth verse having a coin itself. So you have 1 dime you have 10 cents. You have one quarter you ahev 25 cents. So a dime and a quarter yes you have two coins but they are actually worth a 10 and plus that 25, 35 cents. So making sure you designate between denomination and coin is really really important. So we have x times but it worth 10 cents, if I have 2 dimes how much do I have 20 cents, 10 dimes a dollar. If you take the denomination times the amount that you have, so point 1x is the amount of money you have in [IB] same idea for quarters you have 4 quarters you only have 4 pieces of change but you actually have 4 times 25 cents, a dollar. So you want to make sure again you multiply your denomination money times your number of coins you have. And we want these to equal $3.
So I thought just to make sure you understand what we have this is the amount of money we have in dimes plus the amount of money we have in quarters is equal to the total amount of money that we have. From here linear equation 0.1x stays the same distribute this 0.25 in just to make sure I don't do math right 0.25 times 15 is 3.75-0.25x=3. Combine like terms so 0.1x and -0.15x, -0.15x+375=3. Get our x term by itself so subtract over the 375 and then put -0.15x=-0.75 solve it up, divide by that coefficient 0.75 divided by 0.15 is equal to 5, x=5. Okay as always with word problems make sure you're answering the question that you're asked, so the questioning is asking for how many of each denomination do I have?
What we found is x which we said is the number of dimes. So we know that we have 5 dimes we know we have 15 coins total so 5 of those are dimes that would leaves us with 10 quarters. You can always check your work, you have 10 quarters that's 250. 5 dimes is 50 cents 250+50 is $3. So we made our word problem a diagram to an equation and solved it up.
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