Algebra 2 Homework Help!?

| p - 5 | = 2p / 3

p - 5 = 2p / 3

3p - 15 = 2p

p = 15

p - 5 = - 2p / 3

3p - 15 = - 2p

5p = 15

p = 3

Solutions are p = 15 , p = 3

Check

p*******LHS******RHS

15******30********30

(- 2)****3| - 2 |*****6

*********3 x 2******6

*********6**********6

After trying to solve problem after problem on yahooanswers, I honestly think most people dont learn a lot if I give them the answers. Heres a VERY similar problem. Just fill in the values from your own problem. (Hint: you will need to use the Quadratic Formula at the end!) Goodluck!!! QUESTION: The hypotenuse of a right triangle has a length of 13 cm. The sum of the lengths of the two legs is 17 cm. Find the lengths of the legs. Let a = length of one leg Let b = length of other leg We are told that: a + b = 10.23 --> b = 10.23 - b [1] The hypotenuse is 8.312. From Pythagorus: a² + b² = 8.312² [2] Substitute [1] into [2]: a² + (8.312 - b)² = 169 This simplifies to: 2a² - 34a + 120 = 0 Divide by 2: a² - 17a + 60 = 0 which factors: (a - 5)(a - 12) = 0 and has roots: a = 5, 12 Substitute into [1] and get: b = 12, 5 The lengths of the legs are 5 and 12 cm.

3|p-5|=2p

|p-5|=(2/3)p

Remember, the absolute value is always positive and measures the DISTANCE from the origin, that, is what happens to be positive.

(2/3)p can be either positive or negative, so we have to set up two equations, with two answers.

p-5 = (2/3)p ....(1)

(1/3)p = 5

p=15

p-5 = -(2/3)p .... (2)

5/3p = 5

p=3

Thus, p=15, or p=3.

so you get this |p-5|=(2/3)p

since it is absolute value the |p-5| could be either positive or negative (2\3)p. write to separate equations and solve.

p-5= (2/3)p -5=-(1\3)p p=15

p-5=-(2/3)p -5=-(5/3)p p=3

divide 3 for both sides

lp - 5l = 3p/2

l x l = x for x >/ 0 and -x \< 0

use the rule above and you have:

+/- (p - 5) = 3p/2

p - 5 = 2p/3

multiply 2 for both sides

3p - 15 = 2p

subtract 2p for both sides

-p = -15

divide -1 for both sides

p = 15

-(p - 5) = 2p/3

distribute

-p + 5 = 2p/3

multiply 3 for both sides

-3p + 15 = 2p

add 3p for both sides

5p = 15

divide 5 for both sides

p = 3

so p = 3 or 15

after you divide by 3 you get

|p-5| = (2p)/3

so then you need to set up 2 new equations

p-5 = (2p)/3

AND

p-5 = -(2p)\3

add 5 to both sides in both equations then you have your answer

3|p-5| = 2p

|p-5| = 2p/3

then two things can be possible:

p-5 = 2p/3 or p-5 = -2p/3

then you work from there and get two possible answers for p

p= 15 or 3

remember : |p| = a means p = a or p = -a

with this knowledge now you can work on mod functions ....

cheers