This is the kind of Math problem that I promise you're either going to see in your homework, or on your test when it comes to systems of equations. It's going to be almost exactly like this only maybe with different number values.
A jar contains quarters and nickels. There are 15 more nickels that quarter. The total value is $2.55. Find how many of each coin type there are. Okay I'm going to want 2 equations with two variables. My first equation is going to come from this sentence about how many of each coin type there are. There are 15 more nickels than quarters. A lot of students are going to write this 15 more nickels than quarters. There's actually an error here.
What this would mean is that I have to take my n number and add 15 into it in order to get my quarters. That would mean that quarters was the higher number, but that's not true. There are more nickels than there are quarters think about this stuff logically when you're writing these equations. Instead of 15+n=q, I need to do 15 plus q equals n. Now I know I take my quarter’s number and add 15 to it, it has to get larger in order to hit the Ns, the number of nickels.
Okay my second equation is going to come from this second sentence here. The total value is $2.55 cents. In order to deal with value of coins, I'm going to have to use some decimals. .25 represents how much a quarter is worth, plus .05 which represents how much a nickel is worth is equal to $2.55 cents.
If you try a problem like this and you get it incorrect, check and make sure you used .05 for how much a nickel is worth. A lot of students will write it like this. A nickel is 5 cents, incorrect; you need that 0 to show it's .05.
Okay, once I'm at this step, this system of equations is like crying out for substitution because n is already isolated. I can substitute this whole expression 15+q right there. Where it used to be n, it's now going to say 15 plus q. So now I'm going to have only one equation with only one variable and that makes it a lot easier to solve. 2.55 go through and distribute now I'm just simplifying.
.05 times 15, .05 times 15, .75 plus point 05q, I still haven’t done any solving, I'm still just simplifying the left side. Combine your quarter terms and I'll have .3q plus .75 equals 2.55. Subtract .75 from both sides and I have .3q is equal to, here it comes wait for it, wait for it, I know it's exciting drum roll 1. 8. I'm being kind of silly because this is a routine solving problem; you guys have been doing this all year long. I know you can do it. Divide both sides by point 3, and you get that q is equal to 6. I have 6 quarters in that jar.
Once you have that done, you need to find how many nickels also because it acts about both coins types, so in my first equation here, I'm going to substitute 6 for q. N equals 15 plus 6, so n is equal to 21. What that tells me is I have 21 nickels, and 6 quarters in the jar. The best thing to do before you move on is to check your work, make sure that the value of 21 nickels, plus the value of 6 quarters is indeed $2.55.
So when you come to these problems in your homework, the key is to write 2 equations. One equation will be about the amount of each coin type you have, and the other equation will be about the value of those coin types.
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