Question 1165135: The value of the dimes and quarters was $5.75 and there were 5 more dimes than quarters. How many coins of each type were there?
D+Q=5.75 Found 3 solutions by Alan3354, greenestamps, ikleyn:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The value of the dimes and quarters was $5.75 and there were 5 more dimes than quarters. How many coins of each type were there?
D+Q=5.75
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10d + 25q = 575
d = q + 5
Beginning algebra students often simply try writing equations using the numbers given in the problem, without understanding what the equations mean.
In your equation "D+Q=5.75", I presume D and Q are supposed to be the numbers of dimes and quarters. If that is the case, then your equation says that the total number of dimes and quarters is 5.75. That of course is not what the problem says.
It will help you, when you are first learning to solve problems using algebra, to take the time to write out the definitions of the variables and expressions you are going to use.
let D = number of dimes
let Q = number of quarters
Having those definitions written on the paper in front of you will help you write equations that make sense.
One piece of given information is that the number of dimes is 5 more than the number of quarters. Translate that directly into an equation using the variables you have defined:
D = Q+5
The "5.75" (dollars, = 575 cents) in the problem is the VALUE of the dimes and quarters, at 10 cents per dime and 25 cents per quarter:
10D+25Q = 575
Now you have two equations in D and Q which you can solve to find the answer to the problem.
You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.
A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.
Read them and become an expert in solution of coin problems.
