Question

A man bought 42 stamps, some 13¢ and some 18¢. How many of each kind did he buy if the cost was $6.66?

Answer 1

Let's call the $0.13 stamps T and the $0.18 stamps E (for thirteen and eighteen).

T*.13 + E * .18 = 6.66
and
T + E = 42

You can solve the second equation for one of the variables and then substitute;
T = 42- E

Then the first equation becomes
(42-E)*.13 + E * .18 = 6.66

Answer 2

so i got 21 of each kind?

Answer 3

21 * .13 = 2.73
21 * .18 = 3.78

3.78+2.73 = 6.51...So I think we're a little off.

Answer 4

5.46 - E*.13 + E*.18 = 6.66

Subtract 5.46 from both sides to get
0 - E*.13 + E*.18 = 1.20

A little rearranging and distribution property:
E * (.18 - .13) = 1.20
E * (.05) = 1.20

Answer 5

That should be right. I switched the variables in my head. Did that make sense? What did you get?

Answer 6

Systems of Equations:
x + y = 42
.13x + .18 y = 6.66

1. Solve each for y:

y = 42 -x
y = (6.66 - .13x)/.18

2. Set both equations equal to each other:

y = y
42 - x = (6.66 - .13x)/.18
.18(42 - x) = 6.66 - .13x
7.56 - .18x = 6.66 - .13x
.90 = .05x
x = 18
y = 24