# Thread: Got a math problem I'm trying to figure out and need help

1. Member
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## Got a math problem I'm trying to figure out and need help

When makeing the chart, how do you determine which one in the number of coins graph gets "x" and which one gets "#-x"? Such as: Anthony has a total of 30 dimes and quarters worth \$5.70. How many of each type of coin does he have? Do I make Quarters=30-x and Dimes=x or Dimes=30-x and quarters=x? I don't know how to determine which way. Also I get different answers depending one which way I put it. Thanks for any help on this.

Here is an example of one I am saying:

2. 18 Qs and 12 Ds

3. Did you solve the problem both ways?

I think if you do, you will find the answer to your question.

4. Both ways are correct and you will get different answers. The top equation your solving x for # of dimes which is 12. The bottom equation your solving x for # of quarters whish is 18. Either way there are 12 dimes and 18 quarters.

5. Scotty has given you the answer which you would have found if you had solved either of your equations. In fact, if you had solved both equation independently, you would have proved your answer; at least that's the way they taught it when I was in school. Remember, x is an variable in each equation but not the same variable in both equations (try x for dimes and y for quarters and see if that works for you - two equations and two unknowns).

Just to get this thread off on a tangent, please let us know what grade you're in and where you go to school. :wrench in the works:

6. the equation is a binomial because it has 2 dif factors. you would have
to do something like .25x + .10y = 5.70

i am not a math wiz by any stretch but i think this is how to do it.

7. 12 dimes & 18 quarters

8. yea, once again we know the answer, the question was how to state the problem in a equation.

9. Oh shucks! I just don't know how to math with my ABC's.

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Originally Posted by ndzied1
Did you solve the problem both ways?

I think if you do, you will find the answer to your question.
When I solve it the top way I get 12 for the value of x. I know how to solve it from there. When I solve it the bottom way I get 7.71428...........

What I want to know is how do I decide which way to graph my question. I know by solving it both ways and only one works, but why? In the lessons when they do examples they always get it correct the first time, but never say why they chose which way to put it.

Here is an example video:

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Originally Posted by KennB
Just to get this thread off on a tangent, please let us know what grade you're in and where you go to school. :wrench in the works:
I'm not in school right now. I have been out for quite a while and I plan on going to a local college in fall 2012. I just wanted to touch up on things before fall and I have been going thru the lessons at youteacher.com.

12. I get the same answer solving it either way so really, it doesn't matter which way you pick. I think you must have made a mistake when you tried the 2nd way.

13. For this problem, I'd keep Q's and D's as different variables. While using X and 30-X is a valid relationship to describe the amount of Q's and D's, you are using a single variable (X) to represent two different things. You need two equations to solve 2 variables. They are:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)

Rearrange the 2nd equation to isolate a variable.

D=30-Q

I would use substitution and substitute the 2nd equation into the first equation, then start simplifying to find your answer.

.10(30-Q)+.25Q=5.70
3-.10Q+.25Q=5.70
.15Q=2.70
Q= 18

Then you can plug in Q as 18 into either equation to get D's. This will also work if you find D's first. Just rearrange the 2nd equation to isolate Q, and do the same process. There is another way to solve, which is adding the equations. It is set up like this, again starting with the original equations:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)

Next, multiply the entire first equation by -10. You will see why I choose this number in a few seconds. Now the two equations are:

-D-2.5Q=-57
D+Q=30

-D-2.5Q=-57
D +Q=30
-1.5Q = -27
Q = 18

The trick here is to multiply either equation by a number, so that when you add them, a you will get 0 amount of a variable, and they cancel each other out, leaving one variable. So in this case I could find D first by multiplying the first equation by -4, to get a Q and a -Q.
Last edited by COrocket; 19th December 2011 at 10:57 PM.

14. ## Way to go!!

Originally Posted by brandonppr
I'm not in school right now. I have been out for quite a while and I plan on going to a local college in fall 2012. I just wanted to touch up on things before fall and I have been going thru the lessons at youteacher.com.
Brandon,
Congratulations on your plans to attend a local college and in getting your preparation work done this far ahead of time. Check your math on the second equation; the set-up is right and COrocket steps through it for you.

The Your Teacher videos look like a good tool for your prep. It's not the exact method I would use (or was taught) but the logic is workable and the results are correct. You can't argue with that.

Let us know if you'd like any more help working through any other problems. As you've seen, you'll get answers (if that's what you want) and you'll get tutoring (if that's what you want); most of the time you'll get both no matter what you want. For lots of us, it's just nice to share what we've learned and, maybe, haven't had to use so much these days.

I wish you well.
Kenn

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I'm too slow. I was going to type up what the above said, but yeah, what they said.

It should matter what you pick as the X, as long as you remember what it was when you go to substitute it back in to get the answer.

16. Member
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Originally Posted by ndzied1
I get the same answer solving it either way so really, it doesn't matter which way you pick. I think you must have made a mistake when you tried the 2nd way.
Doh! That's exactly what I did wrong. I just re did it and got x=18 for the bottom one. For some reason when I was simplifying (300-10x+25x) I was getting (300-35x) and it should be (300-15x). I guess it was just a brain fart I was doing that on every problem. I see either the top or the bottom equation works now.

Thanks for the help everyone.

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Originally Posted by COrocket
For this problem, I'd keep Q's and D's as different variables. While using X and 30-X is a valid relationship to describe the amount of Q's and D's, you are using a single variable (X) to represent two different things. You need two equations to solve 2 variables. They are:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)

Rearrange the 2nd equation to isolate a variable.

D=30-Q

I would use substitution and substitute the 2nd equation into the first equation, then start simplifying to find your answer.

.10(30-Q)+.25Q=5.70
3-.10Q+.25Q=5.70
.15Q=2.70
Q= 18

Then you can plug in Q as 18 into either equation to get D's. This will also work if you find D's first. Just rearrange the 2nd equation to isolate Q, and do the same process. There is another way to solve, which is adding the equations. It is set up like this, again starting with the original equations:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)

Next, multiply the entire first equation by -10. You will see why I choose this number in a few seconds. Now the two equations are:

-D-2.5Q=-57
D+Q=30

-D-2.5Q=-57
D +Q=30
-1.5Q = -27
Q = 18

The trick here is to multiply either equation by a number, so that when you add them, a you will get 0 amount of a variable, and they cancel each other out, leaving one variable. So in this case I could find D first by multiplying the first equation by -4, to get a Q and a -Q.
OK this is going to take me a while to go over, but I just spoke to my grandfather and he told me that this is how he would solve the problem as well.

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Originally Posted by KennB
Brandon,
Congratulations on your plans to attend a local college and in getting your preparation work done this far ahead of time. Check your math on the second equation; the set-up is right and COrocket steps through it for you.

The Your Teacher videos look like a good tool for your prep. It's not the exact method I would use (or was taught) but the logic is workable and the results are correct. You can't argue with that.

Let us know if you'd like any more help working through any other problems. As you've seen, you'll get answers (if that's what you want) and you'll get tutoring (if that's what you want); most of the time you'll get both no matter what you want. For lots of us, it's just nice to share what we've learned and, maybe, haven't had to use so much these days.

I wish you well.
Kenn
I appreciate the help from everyone.
I am excited to go. I didn't care so much about learning back when I was in highschool, but now I really enjoy learning about how the universe works.

19. Member
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Originally Posted by COrocket
For this problem, I'd keep Q's and D's as different variables. While using X and 30-X is a valid relationship to describe the amount of Q's and D's, you are using a single variable (X) to represent two different things. You need two equations to solve 2 variables. They are:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)

Rearrange the 2nd equation to isolate a variable.

D=30-Q

I would use substitution and substitute the 2nd equation into the first equation, then start simplifying to find your answer.

.10(30-Q)+.25Q=5.70
3-.10Q+.25Q=5.70
.15Q=2.70
Q= 18

Then you can plug in Q as 18 into either equation to get D's. This will also work if you find D's first. Just rearrange the 2nd equation to isolate Q, and do the same process. There is another way to solve, which is adding the equations. It is set up like this, again starting with the original equations:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)

Next, multiply the entire first equation by -10. You will see why I choose this number in a few seconds. Now the two equations are:

-D-2.5Q=-57
D+Q=30

-D-2.5Q=-57
D +Q=30
-1.5Q = -27
Q = 18

The trick here is to multiply either equation by a number, so that when you add them, a you will get 0 amount of a variable, and they cancel each other out, leaving one variable. So in this case I could find D first by multiplying the first equation by -4, to get a Q and a -Q.
Ok I understand this now. Also found a video that teaches it this way: http://www.brightstorm.com/math/alge...ems-problem-1/

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Using that method, how would I write this equation?

Steph has 3 more quarters than nickels. The value of her quarters is 5 cents less than 9 times the value of her nickels. How many of each coin does she have?

21. Originally Posted by brandonppr
Ok I understand this now. Also found a video that teaches it this way: http://www.brightstorm.com/math/alge...ems-problem-1/
Originally Posted by brandonppr
Using that method, how would I write this equation?

Steph has 3 more quarters than nickels. The value of her quarters is 5 cents less than 9 times the value of her nickels. How many of each coin does she have?
Okay, use Problem 2 from the video as that is structured more like this problem. As she states, you need two equations with your two unknowns.

The first equation will be Q = N + 3 to show there are three more quarters than nickels.

The second equation is a little different than the video as you don't know the total value but only a differential value. Your second equation will be .25 x Q = 9(.05 x N) - .05 which can be simplified to 25 x Q = 9(5 x N) - 5 by multiplying each side of the equation by 100 to get rid of the decimals (she stated her solution in decimal form but you may as well work in cents as that's how your example is stated).

Substitute, multiply through and simplify. You should be able to come up with the solution and check your answer. Let us know that you understand how the equations were developed and what you get for your answers. And please keep your eyes on your own paper.
Kenn

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Originally Posted by COrocket
For this problem, I'd keep Q's and D's as different variables. While using X and 30-X is a valid relationship to describe the amount of Q's and D's, you are using a single variable (X) to represent two different things. You need two equations to solve 2 variables. They are:

.10D+.25Q=5.70 (value relationship)
D+Q=30 (quantity relationship)
Can I marry you??

Brandon: think this way, and Matrix Algebra will be a piece of cake. Seriously, arranging terms like this on my Engineering exams, type in the mess and use RREF() on my TI-89, I would blow through problems left and right. It may be easy now with two variables, but how 3x3, or a 5x5 problem (five variables, five equations). As a chem-e we often got several questions of the 5x5 or 6x6 variety on exams, and we only had 40 minutes for the exam most times. It can take 15 minutes to get your 5 or 6 equations right, why spend time hammering out an answer when the ASIC on your little hand computer can belt you out a solution in a second or two?

(love my TI-89, most used instrument on my desk)

I had to go find an article that better describes it, see this:

http://en.wikipedia.org/wiki/System_of_linear_equations
Last edited by darisd; 20th December 2011 at 08:31 PM.

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