pre-algebra problem

• August 20th 2009, 05:12 PM
splits1
pre-algebra problem
Tina, Dawn & Harry have \$175 together. Tina has 3x as much as Dawn. Dawn has 2x as much money as Harry. How much money does each have?
• August 20th 2009, 05:31 PM
Chris L T521
Quote:

Originally Posted by splits1
Tina, Dawn & Harry have \$175 together. Tina has 3x as much as Dawn. Dawn has 2x as much money as Harry. How much money does each have?

Let $T,\,D,\,H$ represent the amount of money Tina, Dawn, and Harry have respectively.

So we know that

$T+D+H=175$ -- (1)

$T=3D$ -- (2)

$D=2H$ -- (3)

Subbing (3) into (2), we can see that

$T=6H$ -- (4)

Now substitute (4) and (3) into (1) to get

$6H+2H+H=175\implies 9H=175\implies H=19.44$

Thus, $T=6(19.44)=116.64$ and $D=2(19.44)=38.88$

(Note that when we go to check our answer, we have $19.44+38.88+116.64=174.96\approx175$ due to rounding errors.)

Does this make sense?
• August 20th 2009, 05:34 PM
QM deFuturo
Are you sure you have copied the problem exactly? Although I can solve this, the end result does not divide evenly. If they had 175.50 together, then each would have an evenly dividable amount.

But, going with what you wrote:

Let x = Harry's money
2x = Dawn's money (since Dawn has 2 times as much money as Harry)
3(2x) = Tina's money (Tina has 3 times as much as Dawn)

All together they have \$175.

x + 2x + 3(2x) = 175

Collect like terms on the LHS, then divide both sides by the coefficient of x. That will solve for x, which is Harry's money. Then multiply x by the right amounts to calculate Dawn and Tina's money.
• August 20th 2009, 05:35 PM
splits1
Yes, it makes sense. Thank you very much!!