Math Homework Need Help!?

1) Use B for Brianne and C for Camila. Right now:

B = 3C (B is 3 times older than C)

In 4 years, B will be twice C's age:

B + 4 = 2 (C + 4) = 2C + 8

Now you can just subtract the the first formula from the second formula:

B + 4 = 2C + 8

- B = 3C

= 4 = -C + 8

Subtract 4 from and add C to both sides:

4 - 4 + C = -C + 8 - 4 + C

C = 4

So Camila is 4 and Brianne (who is currently 3 times C's age) is 12

2) In this case, order does not matter, you just want 3 unique players. That means a team of players A B C is the same as B A C and C B A. This is called a combination. The formula is C(n, k) = n! / (k! (n - k)!) where n = 8 and k = 3

C(8,3) = 8! / 3!(8-3)!

C(8,3) = (8x7x6x5x4x3x2x1) / ((3x2x1)x(5x4x3x2x1)

Cancel out like terms from numerator and denominator, which are 5, 4, 3, 2, 1 to get...

C(8,3) = 8x7x6 / 3x2x1

Notice that the denominator is 6, which appears in the numerator, so cancel those out to get...

C(8,3) = 8x7 = 56

So there are 45 possible different teams of 3 you can form from a group of 8 players.

Here's the first one:

Let B = Brianne's present age and C = Camila's present age.

(1) B = 3C

(2) B + 4 = 2(C + 4) substitute (1) into (2):

3C + 4 = 2(C + 4)

3C + 4 = 2C + 8

C = 4 years old; now substitute back into (1):

B = 3(4) = 12 years old

check using (2):

12 + 4 = 2(4 + 4)