Question

3. After a PTA Fun Night at school, the Water Fountain game had many coins at the
bottom. When the coins were counted, it was found that the number of dimes was
half the number of quarters, the number of nickels was 7 more than three times the
number of quarters, and there were 34 pennies. If the total in the foutain was $18.69.
how many of each coin were in the fountain

Answer 1

Someone please help!

Answer 2

four types of coins. we will represent them as q for quarters, d for dimes, n for nickels, and p for pennies

translating the problem into equation form:

"number of dimes was half the number of quarters" ----> d = (1/2)q
"the number of nickels was 7 more than three times the number of quarters" ---> n = 3q + 7
"and there were 34 pennies" ---> p = 34

so far, our system is:

d = (1/2)q
n = 3q + 7
p = 34

Answer 3

now, we can also write one more equation, that expresses the total amount of money in terms of the # of coins

the total amount is $18.69, made up of quarters, pennies, nickels, dimes

a quarter is 25 cents, so the total value of all the quarters is 0.25q
similar logic with the other coins: 0.1d, 0.05n, 0.01p

summing these up:

0.25q + 0.1d + 0.05n + 0.01p = $18.69

Answer 4

so what we have so far:
0.25q + 0.1d + 0.05n + 0.01p = $18.69
d = (1/2)q
n = 3q + 7
p = 34

notice how if you plug in the d, n, and p equations into the first big equation, you'll eliminate all variables except q. therefore, you can solve for q. from there, you can determine how many dimes and quarters there are using the d = and n = equations.