Question 151661This question is from textbook elementary and intermediate algebra concepts and applications
: Can someone please help me im stuck?
9. Sandy has $3.75 in change. She has three more quarters than dimes but twice as many nickels as quarters. Express the given information as an algebraic expression and simplify. [You are not required to solve the problem]
This question is from textbook elementary and intermediate algebra concepts and applications
Answer by mducky2(62)   (Show Source):
You can put this solution on YOUR website! In order to set up the equation, you must multiply the value of each coin times the number of each coin and add it up. Although we know the value of each coin, we only know the proportions of the amount of each coin. We can label the number of coins:
Number of quarters: q
Number of dimes: d
Number of nickels: n
Multiplying the values of each should come out to be:
25q + 10d + 5n = 375
However, we can't solve this equation until we only have one variable.
Because we know the proportions, we can solve for ratios of each coin in terms of each other. Both ratios include quarters, so let's solve for the amounts in terms proportion. Since there are three more quarters than dimes, we can solve for dimes in terms of quarters:
q = d + 3
d = q - 3
Since there are two times as many nickels as there are dimes, we can set up the equation:
n = 2q
Plugging this into the original equation above:
25q + 10d + 5n = 375
25q + 10(q-3) + 5(2q) = 375
25q + 10q - 30 + 10q = 375
45q = 375 + 30
45q = 405
q = 9
Sandy has nine quarters.
Now we can plug that into the other equations:
d = q - 3
d = 9 - 3
d = 6
n = 2q
n = 2(9)
n = 18
Therefore, Sandy has 9 quarters, 6 dimes, and 18 nickels.
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