SOLUTION: You have 23 coins, including nickels, dimes, and quarters. If you have two more dimes than quarters, and the total value of the coins is $2.50. How many of each kind of coin do y
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Question 1114360: You have 23 coins, including nickels, dimes, and quarters. If you have two more dimes than quarters, and the total value of the coins is $2.50. How many of each kind of coin do you have?
Define the variables you will be using to solve this problem.
Set up the problem by using algebraic equations to represent the problem.
Put the equations in matrix form.
q+d+n=23
D=q+2
0.25q+0.10d+0.05n=$2.50
0.25q+0.10(q+2)+0.05n=2.50
25q+10q+5n=250
35q+5n=250
I am so confused with this problem, I started to set it up and blanked. Am I even close to the algebraic equation?
Your first equation is
Q + D + N = 23
Your second equation is
D = Q + 2.
From these two equations you have
Q + (Q+2) + N = 23
2Q + N = 21. (1)
Now the money equation is
25Q + 10D + 5N = 250 cents, or
25Q + 10*(Q+2) + 5N = 250, or, equivalently
35Q + 5N = 230. (2)
Thus you have this system of two equations in two unknowns
2Q + N = 21, (1)
35Q + 5N = 230 (2)
Divide eq(2) by 5 (both sides). Keep eq(1) as is.
2Q + N = 21 (1')
7Q + N = 46 (2')
Now subtract eq(1') from eq(2')
5Q = 46 - 21 = 25 ====> Q = = 5.
5 quarters. From (1') N = 21 - 2Q = 21 - 2*5 = 11.
11 nickels. Dimes = Q+2 = 5+2 = 7.
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and for all methods used - from one equation to two equations and even without equations.
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