Word Problem Help?

Hey I need some help setting up these two algebra word probelms:



A combination of quarters and dimes equals $1.65. If there are 6 more dimes than there are quarters, how many coins of each are there?



And:



Roses are sold for $3.50 per dozen and carnation are sold for $2.50 per dozen. If the florist sold 14 dozen and $43, how many dozen of each flower did the florist sell?



Thanks. :-)

Word Problem Help?
Let q = number of quarters

Let d = number of dimes



0.25 * q = the worth of the quarters

0.10 * d = the worth of the dimes



0.25 * q + 0.10*d = the total worth of quarters and dimes



0.25 * q + 0.10*d = 1.65 %26lt;========= from problem



6 more dimes than quarters means you have to multiply the number of QUARTERS by 6 to equal the number of dimes



6q = d %26lt;=== OOPS! I READ THE PROBLEM AS 6 TIMES MORE AND NOT 6 MORE. This should be d - 6 = q

SO SORRY FOR THE MISTAKE. If I would have tried to solve it, it probably would have become evident that my equation was wrong. . . . . . .



There, it's set up -- just make a substitution in the line above to solve for q %26amp; d. . .



---------------------------------

3.5r + 2.5c = 43 %26lt;== total cost of (r)oses %26amp; (c)arnations

r + c = 14 %26lt;==== total number of (r)oses %26amp; (c)arnation (in dozens)
Reply:quantity equation

d = q + 6

value equation

10d + 25q = 165



Substitute

10(q + 6) + 25q = 165

10q + 60 + 25q = 165

35q = 105

q = 3



d = q + 6 = 9



9 dimes %26amp; 3 quarters will have a value of

90￠ + 75￠ = $1.65



quantity equation

r + c = 14

r = 14 - c

value equation

3.5r + 2.5c = 43



Substitute

3.5(14 - c) + 2.5c = 43

49 - 3.5c + 2.5c = 43

-c = -6

c = 6



r = 14 - c = 8



8 dozen roses %26amp; 6 dozen carnations will cost

8($3.50) + 6($2.50) = $28.00 + $15.00 = $43.00
Reply:1. let q = quarters and d = dimes

q + 6= d

.25q + .10d = 1.65



Now just use subsitution:

.25q + .10(q + 6) = 1.65

.25q +.10q +.60 = 1.65

.35q = 1.05

q = 3

If there are 3 quarters, then there are 9 dimes.

Check: 3 quarters = .75 and 9 dimes = .90

That adds up to 1.65.



2. Let r = roses and c = carnations

r + c = 14

3.50r + 2.50c = 43

Again, use substitution, or you can add/subtract the two equations, it's up to you. I will use substitution, using the equation r = 14 - c

3.50(14 - c) + 2.50c = 43

49 - 3.50c + 2.50c = 43

-1.00c = -6

c = 6

The florist sold 6 dozen carnations and 8 dozen roses.