Question 120637This question is from textbook Algebra Structure and Method
: I have been trying for two hours to get this, but I can't. Can someone PLEASE help???
Solve by using three equations and variables.
I have 30 coins, all nickels, dimes, and quarters, worth $4.60. There are two more dimes than quarters. How many of each kind of coin do I have?
This question is from textbook Algebra Structure and Method
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! I have 30 coins, all nickels, dimes, and quarters, worth $4.60. There are two more dimes than quarters. How many of each kind of coin do I have?
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Let: n = no. of nickels, d = no. of dime, q = no. of quarters
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Write an equation for each phrase:
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I have 30 coins,
Eq1: n + d + q = 30
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all nickels, dimes, and quarters, worth $4.60.
Eq2: .05n + .10d + .25q = 4.60
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There are two more dimes than quarters.
d = q + 2
or
eq3: d - q = 2
:
How many of each kind of coin do I have?
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Use the 1st and 3rd equations to eliminate q
n + d + q = 30
0 + d - q = 2
---------------- adding eliminates q
n + 2d +0 = 32;
n + 2d = 32; our 1st "two unknown" equation
:
Multiply eq2 by 4 and subtract it from eq1
1n + 1d + 1q = 30.00
.2n+ .4d + 1q =18.40
-----------------------subtracting eliminates q again:
.8n + .6d + 0 = 11.60
.8n + .6d = 11.6; our 2nd "two unknown equation
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Multiply the 1st "two unknown equation by .3 and subtract it from above equation
.8n + .6d = 11.6
.3n + .6d = 9.6
--------------------subtracting eliminates d, find n:
.5n + 0d = 2
n = 2/.5
n = 4 nickels
:
Use the 1st "two unknown" equation to find d
n + 2d = 32
4 + 2d = 32
2d = 32 - 4
d = 28/2
d = 14 dimes
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Use the eq1 to find q:
4 + 14 + q = 30
q = 30 - 18
q = 12 quarters
:
Check solutions in the $value equation
.05(4) + .10(14) + .25(12) =
.20 + 1.40 + 3.00 = 4.60 as given
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