- Thread starter sams
- Start date

- Joined
- Feb 4, 2004

- Messages
- 15,945

If you have one quarter, how many cents is that? If you have two, how many cents is that? What about three? Four? How about if you have "q" quarters?

Follow this reasoning to find an expression for the value represented by q quarters, and a similar expression for the value represented by h half-dollars.

If you have one quarter, how many of the sixteen coins are half-dollars? What if you have two quarters? Three? Four? How about if you have "q" quarters?

Follow this reasoning to express "h" in terms of "q". Then plug this expression in for "h" in your "value of the half-dollars" expression. This gives you expressions for the values of each coins, all in terms of the one variable.

Sum these two expressions, and set equal to "625", the total value (in cents). Solve.

Eliz.

- Joined
- Apr 12, 2005

- Messages
- 9,990

You should not expect this to work.sams said:.25x + .50x = 6.25 (doesn't work)

.50x + 2(.25x) = 16 (doesn't work)

1) You have provided no definition for 'x'. If you solve for it, what will you have?

2) You have used 'x' for both # of quarters and # of halves. The problem statement does not say the counts are the same.

This is better, but it remains an insufficient definition. Be very specific and explicit. If you had provided the exact problem statement, a better definition would be possible.if x is quarters, y is half dollars

x + y = 16

.25x + .50y = 6.25

Can you solve this system of two variables in two equations?

"There is a collection of coins of half dollars and quarters. The total number of coins is 16 and the value of the coins is $6.25."

My daughter has several problems like this on her algebra homework. Unfortunately, she was absent when the teacher went over the problems and has yet to ask the teacher for the proper solution....

- Joined
- Feb 4, 2004

- Messages
- 15,945

So you haven't yet read the outline of how to solve this algebraically, in the first reply (above) to this thread...?sams said:To date, I still do not have a solution to this problem other than to do it trial and error.

Eliz.

You have the right system of equations there.sams said:x + y = 16

.25x + .50y = 6.25 ???

x+y = 16

0.25x+0.50y = 6.25

2(0.25x+0.50y) = 2(6.25)

0.5x+y = 12.5

Solve it by substitution or subtraction.