Question 1175437: A child is collecting state quarters and new $1 coins. If she has a total of 21 coins, and the number of quarters is 9 more than the number of dollar coins, how many of each type of coin does she have?
Found 2 solutions by ikleyn, ankor@dixie-net.com:
Answer by ikleyn(52756)   (Show Source):
You can put this solution on YOUR website! .
You can solve it in this way, using one single equation
D + (D + 9) = 21 coins
2D = 21 - 9 = 12
D = 12/2 = 6
obtaining the answer 6 one-dollar coins and 6+9 = 15 quarters.
Or, alternatively, you can write a system of two equations
Q + D = 21
Q - D = 9
and then add the equations, obtaining
2Q = 21 + 9 = 30
Q = 30/2 = 15 quarters and 15-9 = 6 one-dollar coins.
You can choose any of these two way - they lead you to the same answer.
Solved, explained and completed.
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Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! A child is collecting state quarters and new $1 coins.
If she has a total of 21 coins, and the number of quarters is 9 more than the number of dollar coins, how many of each type of coin does she have?
let x = no. of dollar coins
She has 9 more quarter than dollar coins, therefore
(x+9) = no. of quarters
:
x + (x+9) = 21
2x = 21 - 9
2x = 12
x = 12/2
x = 6 dollar coins
:
I'll let you find the number of quarters, see that they add up to 21
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