SOLUTION: A man has twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes he would have 90 cents more than he has now. How many dimes and quar
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Question 139131: A man has twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes he would have 90 cents more than he has now. How many dimes and quarters does he have.
.10x+.25y=z
.25x+.10y=z+.90
I am stuck.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A man has twenty coins consisting of dimes and quarters. If the dimes were quarters and the quarters were dimes he would have 90 cents more than he has now. How many dimes and quarters does he have.
.10x+.25y=z
.25x+.10y=z+.90
You have the right idea, just omitted the total coin equation.
:
the total coin equation:
x + y = 20
or
x = (20 - y); we can use this for substitution
:
Write an equation for the statement:
"If the dimes were quarters and the quarters were dimes he would have 90 cents more than he has now."
.10x + .25y + .90 = .25x + .10y; (you don't need z)
:
Combine the y's on the left and the x's on the right (avoids negatives)
.25y - .10y + .90 = .25x - .10x
.15y + .90 = .15x
:
From the total coins equation, substitute (20-y) for x, solve for y:
.15y + .90 = .15(20-y)
.15y + .90 = 3 - .15y
.15y + .15y = 3 - .90
.30y = 2.10
y =
y = 7 quarters
:
I'll let you find the number of dimes. Check your solutions in the given statement.
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