SOLUTION: G'day, Thanks in advance for taking the time to answer my question. Here it is: A collection of 70 coins consisting of dimes, quarters, and half-dollars has a value of $17

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Question 935847: G'day,
Thanks in advance for taking the time to answer my question.
Here it is:
A collection of 70 coins consisting of dimes, quarters, and half-dollars has a value of $17.75. There are three times as many quarters as dimes. Find the number of each kind of coin.
I'm not even sure how to build the equation for this problem.
Let:
Number of dimes = d
Number of quarters = 3d
Number of half dollars = 0.5(70) - 4d ???
If you could help me write this equation it would be most appreciated.
Thank you so much!
Kind regards,
Mark Beaton
Found 3 solutions by TimothyLamb, josgarithmetic, MathTherapy:
Answer by TimothyLamb(4379)   (Show Source): You can put this solution on YOUR website!
x = number of dimes
y = number of quarters
z = number of half-dollars
---
x + y + z = 70
10x + 25y + 50z = 1775
y = 3x
---
put the system of linear equations into standard form
---
x + y + z = 70
10x + 25y + 50z = 1775
3x - y = 0
---
copy and paste the above standard form linear equations in to this solver:
https://sooeet.com/math/system-of-linear-equations-solver.php
---
solution:
x = number of dimes = 15
y = number of quarters = 45
z = number of half-dollars = 10
---
now by elimination:
---
x + y + z = 70
3x - y = 0
---
add the above two equations:
---
4x + z = 70
---
10x + 25y + 50z = 1775
25(3x - y = 0)
---
add the above two equations:
---
85x + 50z = 1775
---
-50(4x + z = 70)
85x + 50z = 1775
---
add the above two equations:
---
-115x = 1775 - 3500
x = (1775 - 3500)/-115
x = 15
---
3x - y = 0
y = 3*15
y = 45
---
4x + z = 70
4*15 + z = 70
z = 10
---
solution by elimination:
x = number of dimes = 15
y = number of quarters = 45
z = number of half-dollars = 10
---
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Answer by josgarithmetic(39623)   (Show Source): You can put this solution on YOUR website!
d, q, h for dimes quarters half-dollars;
Coin count: ;
Ratio description:;
Money count:

The money count equation,



The ratio equation becomes .

System of Equations:

The easiest thing to do is substitute for q in the money count and coin count equations, and then you have a system of two equations in two unknowns, d and h.


The process of solving

The money count,
, when substituting according to q=3d.



The coin count,
, when substituting q=3d.


You have now a simpler system in d and h:
.
Solve this system using either substitution method or Elimination method. Guessing that you forgot how to use Elimination, solve one of these equations for either variable and substitute this into the other equation, and solve for the value of the single variable present.

Solving first for h seems easiest.


'





when multiplied both sides by .




You should be able to work through finding the values for q and h.
Answer by MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!

G'day,
Thanks in advance for taking the time to answer my question.
Here it is:
A collection of 70 coins consisting of dimes, quarters, and half-dollars has a value of $17.75. There are three times as many quarters as dimes. Find the number of each kind of coin.
I'm not even sure how to build the equation for this problem.
Let:
Number of dimes = d
Number of quarters = 3d
Number of half dollars = 0.5(70) - 4d ???
If you could help me write this equation it would be most appreciated.
Thank you so much!
Kind regards,
Mark Beaton
Let number of dimes, be d

Then number of quarters = 3d

Since there are 70 coins, the number of half-dollars is: 70 - (d + 3d), or 70 - 4d

The following value equation can then be formed: .1d + .25(3d) + .5(70 - 4d) = 17.75

Solve this for d, the number of dimes, then you should be able to determine the number of quarters,

and the number of half-dollars.

You then do a check!! 

===================

If you need a complete and detailed solution, let me know!!

Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.

Further help is available, online or in-person, for a fee, obviously. 



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