Re: Beginning Algebra
sandyk109 said:
Jessica has 16 dimes and quarters. Whitney has twice twice as many dimes and 1/3 as many quarters as Jessica has. If they both have the same amount of money, what coins does each have?
Let's try it using one unknown and see what happens.
First, define what you know.
\(\displaystyle Let \ \ d = no.\ \ dimes \ \ Jessica \ \ has\)
\(\displaystyle Let \ \ 16-d = no.\ \ quarters \ \ Jessica \ \ has\)
\(\displaystyle Let \ \ 2d = no.\ \ dimes \ \ Whitney \ \ has\)
\(\displaystyle Let \ \ \frac{1}{3}(16-d) = no.\ \ quarters \ \ Whitney \ \ has\)
All that came from the first two sentences.
A dime is worth
10 cents. So
d dimes would be worth
10d cents, and
2d dimes would be worth
10(2d) cents.
Similarly, a quarter is worth
25 cents, so
16-d quarters would be worth
25(16-d) cents, and \(\displaystyle \frac{1}{3}(16-d)\) quarters would be worth \(\displaystyle 25(\frac{1}{3})(16-d)\) cents.
Since Jessica and Whitney have the same amount of money, we can build this equation:
\(\displaystyle \overbrace{10d+25(16-d)}^{\text{Jessica's amount}}=\overbrace{10(2d)+25(\frac{1}{3})(16-d)}^{\text{Whitney's amount}}\)
I would probably multiply each term by
3 to eliminate the fraction and then solve from there.
Can you finish from here? Once you solve for
d, substitute that back into the definitions we started with to find out how many dimes and quarters each girl has.
Good luck. Let us know if you were successful.