WORD PROBLEM pleas help

Question

Brandon has a jar of quarters and dimes with a total value of $5.70. The number of quarters is 6 less than twice the number of dimes. How many quarters and how many dimes does Brandon have?

Answer 1

This is a lovely value/total word problem to be solved, in my way of thinking, with a nice little system of two linear equations: one for the value and another from the information in the question relating the quantities of coins. I'll use "q" for number of quarters and "d" for number of dimes.

First the value of all the coins:

.25q + .10d = 5.70

which I will immediately multiply by 100 to get rid of the decimals:

25q + 10d = 570

Then the other piece of information about the relationship between the numbers of coins gives me

q = 2d - 6

Substitute that into the previous equation and you will have solved the system by substitution. Be sure to check your answer!

Answer 2

Let D = the number of Dimes and Let Q = the number of Quarters.

.25Q + .10D = 5.70 and

Q = 2D - 6

Looks like we have to solve for TWO SIMULTANEOUS LINEAR EQUATIONS here.

Multiply everything by 100 to rid ourselves of those decimals.

25Q + 10D = 570

Now, substitute 2D - 6 for "Q" in the first equation.

25(2D - 6) + 10D = 570

50D - 150 + 10D = 570

60D - 150 = 570

60D = 720 by adding 150 to both sides of the equation.

6D = 72 dividing both sides of the equation by 10

D = 12

So, Brandon has 12 dimes which is $1.20.

Q = 2D - 6 from the second equation is the number of quarters.

Q = 2(12) - 6

= 24 - 6

= 18

Brandon has 18 Quarters which is $4.50 . (.25 X 18 = $4.50)

CHECKING OUR ANSWERS:

$1.20

+ $4.50

=========

$5.70

BRANDON HAS 12 DIMES AND 18 QUARTERS