##### 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

--------------------------------------------------------------------------------------------------

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

----------------------------------------------------------------------------------------------------

Jun 19th, 2015

...
Jun 19th, 2015
...
Jun 19th, 2015
Dec 10th, 2016
check_circle