If both have same amt of money, what coins does each have?

[sandyk109]

Jessica has 16 dimes and quarters. Whitney has twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what conis does each have?

10(16 -x) + 25x = 2[10(16-x)] + 1/3 x

3(160 -10x +25x = 2[160- 10x]

480 -30x +75x = 320 -20x

160 = -26X

Please help

 

[Deleted member 4993]

Re: Beginning Algebra

Jessica has 16 dimes and quarters. Whitney has twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what conis does each have?

10(16 -x) + 25x = 2[10(16-x)] + 1/3 x

3(160 -10x +25x = 2[160- 10x]

480 -30x +75x = 320 -20x

160 = -26X

Please help

Assume Jessica has "D" dimes and "16-D" quarters.

How much money does Jessica have?.........................................(1)

How many dimes does Whtiney have ? = 2*D

How many quarters does Whtiney have ?

How much money does Whiney have?.........................................(2)

Equate (1) and (2) - and solve for 'D'.

If you are still stuck - write back showing work.

 

[TchrWill]

Re: Beginning Algebra

Jessica has 16 dimes and quarters. Whitney has twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what conis does each have?

Jessica has D dimes and Q quarters or D + Q
Whitnet has2D dimes and Q/3 quarters.

Therefore, .10D + .25Q = 2(.10)D + .25Q/3
Multiplying through by 300 yields 30D + 75Q = 60D + 25Q or 30D = 50Q making D/Q = 1.6666 or Q/D = .60.

The first integers that satisfy this requirement are 5D and 3 quarters.
Jessica has 5 dimes and 3 quarters for .10(50) + .25(3) = .50 + .75 = $1.25
Whitney has 10 dimes and 1 quarters for .10(10) + .25(12) = 1.00 + .25 = $1.25

But hold on. any multiple of these numbers also gives them the same $ amount. The coins shown above are the minimum amount they could have.