You will need to set up your table of values for each person.
Let x represent the months, and y for the amount of money.
For Carissa: For Louann: x = y x = y -- -- -- -- 0 $250 0 $1230 1 $330 1 $1170 2 $410 2 $1110 3 $490 3 $1050 4 $570 4 $990 5 $650 5 $930 6 $730 6 $870 7 $810 7 $810
You could also do the equation: 80x + 250 = -60x + 1250 where you will get x=7. Then substitute 7 to the x's in the equation which will give you $810 for each.
The answer is: It will take 7 months. Then, they will both have $810 in their accounts.
This exercise is trying to get you to write an equation that describes what's going on.
Carissa has $250 today, and she adds $80 to it each month. In 'm' months from now, she'll have 80m more than $250.
C (for Carissa) = 250 + 80m
Louann has $1230 today, and she takes out $60 each month. In 'm' months from now, she'll have 60m less than $1230.
L (for Louann) = 1230 - 60m
Carissa has less money today, but it's growing $80 every month. Louann has more money today, but it's losing $60 every month. Eventually, they'll both have the same amount. The question is: When and how much ?
Well, when they both have the same amount, then C = L .
250 + 80m = 1230 - 60m
That's the end of the hard part of this problem ... writing the equation. The rest is easy, and I'm sure you'd have no trouble solving it. But since I'm on a roll, and I'm going to take your 5 points, I might as well keep going:
250 + 80m = 1230 - 60m
Subtract 250 from each side: 80m = 980 - 60m
Add 60m to each side: 140m = 980
Divide each side by 140 : m = 7
This is telling us that after 7 months, Carissa and Louann will both have the same amount of money in their accounts.
The problem also asks us how much that is. So let's find the amount for both of them, just to check our work and make sure they're both the same:
Here we have to look at the definition of terminate, which means "bring to an end" a terminating decimal is a decimal that, as previously stated, comes to an end. 0.125 is a decimal that ends, therefore is a terminating decimal.
