Arnold had 1.70 in dimes and quarters. Help with writing an equation from the word problem

Algebra
Tutor: None Selected Time limit: 1 Day

Arnold had $1.70 in dimes and quarters. He had 3 more dimes than quarters. How many of each coin did he have? 

I need help writing the equation from this step by step. 

Jun 19th, 2015

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Let the number of dimes be x

and the number of quarters be y.


Convert dimes, quarters and dollars into cents.

1 dime = 10 cents

So, x dimes = 10x cents

1 quarter = 25 cents

So, y quarters = 25y cents

1 dollar = 100 cents

So, 1.70 dollar = 1.70*100 cents = 170 cents


Arnold had $1.70 in dimes and quarters.

10x + 25y = 170 

Divide throughout by 5

2x + 5y = 34 ---(1)


He had 3 more dimes than quarters.

x = y + 3 ---(2)


Substitute x from equation (2) into equation (1)

2(y + 3) + 5y = 34

2y + 6 + 5y = 34

7y = 34 - 6

7y = 28

y = 28/7

y = 4


Substitute value of y in equation (2)

x = 4 + 3 

x = 7


So, number of dimes = x = 7

and number of quarters = y = 4

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Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 19th, 2015

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