# Word problem w/ three variables

• Oct 19th 2010, 05:26 PM
gustahfan
Word problem w/ three variables
This problem, is seriously bugging me:

A collection of nickels, dimes, and quarters consists of 12 coins with a total value of $1.45. If the number of nickels is 2 less than the number of dimes, how many of each coin are contained in the collection? I thought, I had it right the first time, but my totals didn't add up to 1.45. I don't understand what I am doing wrong. The three equations I came up with are: x + y + z = 12 x = y - 2 or x - y = -2 5x + 10y + 25z = 145 Are my equations wrong? • Oct 19th 2010, 06:06 PM pickslides I don't think your equations are wrong. 'How did you solve them?' is the question on my mind. Because the solutions don't seem to work. • Oct 19th 2010, 06:17 PM gustahfan x + y + z = 12 z - y = -2 5x + 10y + 25z = 145 x + y + z = 12 x - y = -2 2x + z = 10 x + y + z = 12 5x + 10y + 25z = 145 10x + 10y + 10z = 120 5x + 10y + 25z = 145 5x - 15z = -25 2x + z = 10 5z - 15z = -25 30x + 15z = 150 5x - 15z = -25 35x = 125 x = NOT RIGHT! Well originally I had 4 here, but while typing it out, I see that I subtracted and also had 125 instead of 150 on the last step to get 25x = 100, but it should have been addition...and now it's alllllllllll messed up again. Does anyone see an error in my math? • Oct 24th 2010, 04:10 PM gustahfan I even went as far as trying to say that if x = 1 - 12, and y being two more...and not any of those answers were correct. Is something wrong with this problem or me!!! LOL!!! I keep coming up with this... x =$\displaystyle \displaystyle{\frac{25}{7}$y=$\displaystyle \displaystyle{\frac{39}{7}$z=$\displaystyle \displaystyle{\frac{20}{7}$And if this is right I am going to uppercut my teacher... • Oct 24th 2010, 04:46 PM harish21 Quote: Originally Posted by gustahfan BUMP! :) I even went as far as trying to say that if x = 1 - 12, and y being two more...and not any of those answers were correct. Is something wrong with this problem or me!!! LOL!!! I keep coming up with this... x =$\displaystyle \displaystyle{\frac{25}{7}$y=$\displaystyle \displaystyle{\frac{39}{7}$z=$\displaystyle \displaystyle{\frac{20}{7}\$

And if this is right I am going to uppercut my teacher...

The equations you have set up in Post 1 look correct.

And the values that you have mentioned for x,y,and z are the only ones that are going to fit your equations.

Life would have been simpler if the total number of coins was 14.
• Oct 24th 2010, 04:51 PM
gustahfan
Quote:

Originally Posted by harish21
The equations you have set up in Post 1 look correct.

And the values that you have mentioned for x,y,and z are the only ones that are going to fit your equations.

Life would have been simpler if the total number of coins was 14.

This teacher is so cruel to us!
• Oct 24th 2010, 07:14 PM
Wilmer
If you change total coins from 12 to 13, then all's well that ends well...
6 nickels + 4 dimes + 3 quarters = .30 + .40 + .75 = 1.45
• Oct 25th 2010, 09:26 PM
gustahfan
yea, she told us today she made a mistake...you think?? !!