Question

hi

Answer 1

\(q \ne 0\) goes without saying. This is a very stumpable problem. Where'd you find it?

Answer 2

ok, so if \[p-q=p+q=pq\], which is not possible because then p=0 and q=0

Answer 3

so if \[p+q=pq=\frac{ p }{ q }\] then \[p=\frac{1}{2}\] and \[q=-1\] which implies that \[p-q =\frac{ -1 }{2 }\] which is again, impossible

Answer 4

ok got it

Answer 5

this works: \[p-q=pq=\frac{p}{q}\] because this yields \[p=\frac{-1}{2} and q=-1\] so \[p+q=\frac{-3}{2}\]

Answer 6

is this correct?

Answer 7

If it works then you should be able to demonstrate that it works.

Answer 8

final answer \[\frac{3}{2}\]

Answer 9

um, i can, because by demonstrating all four cases and how only ONE of them works, that must be the answer

Answer 10

can you help me with another problem thats giving me more difficulty?

Answer 11

I came up with those numbers before, but I believe I used q = -1/2 and p = -1 instead of what you have done.

Answer 12

I was very close, I just didn't think to swap the p and q

Answer 13

wait one sec, how would one Algebraically solve the system p-q=pq=p/q? i used graphing to do it. how would one solve that algebraically?

Answer 14

You used graphing to do it? Hmmm.

Answer 15

yea, i graphed the 3 systems and found their intersection but i want to be able to solve it algebraically

Answer 16

That's cheating.

Answer 17

I'll have to re-trace my steps for this.

Answer 18

I started with
pq = p/q
q^2 = 1
q = ± 1

Then I used
p - q = pq
p - (-1) = -p
p + 1 = -p
2p = -1
p = -1/2

Answer 19

I used q = -1

Answer 20

okay, got it.

Answer 21

what happens when q=1?

Answer 22

OHHHH that yields the exact same answer

Answer 23

so it doesnt really matter

Answer 24

Well, if that's the case, then you have two solutions.

Answer 25

no, not really. The question asks for the absolute value of the fourth expression (p+q), which is, in either case, 3/2

Answer 26

You have two solutions for q if it works.

Answer 27

oh, for q yes

Answer 28

could you help me with one more?

Answer 29

You can post your question and we'll go from there. It is not necessary to ask for permission to post a question.

Answer 30

ok thanks