SOLUTION: Here is the queston:
how do I set up the equation?
A collection of coins is worth $7.50. There are nickles, dimes and quarters with twice as many dimes as there are nickles. Alto
Algebra.Com
Question 13602: Here is the queston:
how do I set up the equation?
A collection of coins is worth $7.50. There are nickles, dimes and quarters with twice as many dimes as there are nickles. Altogether there are 70 coins. how many nickles, dines, and quarters are in the collection?
Answer by LilSkittleMd(119) (Show Source): You can put this solution on YOUR website!
For this problem you have 3 equations. They are:
n+d+q=70
(since there are 70 coins altogether, the number of each coin will add up to 70)
d=2n
(there are twice as many dimes as there are nickels)
0.05n+0.10d+0.25q=7.50
(the amount of how much each coin is worth, multiplied by the number of the coin is equal to $7.50)
The first step is to eliminate one variable from two equations. Using the elimination method seems to be the easiest (add the two equations together, eliminating one variable)
Take the first and last equation. Multiply the first equation by -0.25 to cancel out the variable q.
-0.25n-0.25d-0.25q=-17.5
+ 0.05n+0.10d+0.25q=7.50 (Add the two equations to come up with an equation without the variable q)
-0.2n-0.15d=-10
Take the d=n equation and plug it into the new equation:
-0.2n-0.15(2n)=-10 (Solve the equation)
-0.2n-0.3n=-10
-0.5n=-10 (divide each side of the equation by -0.5)
n=20
Now place the value of n into an equation:
d=2(20)
d=40
Now place these 2 variables into an equation with all three variables:
20+40+q=70 (Solve this equation)
60+q=70
q=10
There are 20 nickels, 40 dimes, and 10 quarters in the collection.
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