Question

\(\large \tt \begin{align} \color{black}{\dfrac{P+1}{Q+1}=\dfrac{R}{S}\hspace{.33em}\\~\\
\dfrac{P+2}{Q+2}=\dfrac{1}{2}\hspace{.33em}\\~\\
P+Q=?\hspace{.33em}\\~\\
a.3\hspace{.33em}\\~\\
b.4\hspace{.33em}\\~\\
c.5\hspace{.33em}\\~\\
d.6\hspace{.33em}\\~\\
P,Q,R,S \in \mathbb{Z} \hspace{.33em}\\~\\

}\end{align}\)

Answer 1

Are we given any other info? Are these all integers or are we given r and s?

Answer 2

r and s are integers, no other info is given

Answer 3

and p and q also are integers

Answer 4

I am just sort of playing around with it right now, but we can safely say that (P+2) and (Q+2) are either both positive or both negative, which might come in useful.

Answer 5

yes i get they can be both \(\pm\) , i think the option are used to solve this

Answer 6

what is R/s ?

Answer 7

Right, so P and Q are both positive, R and S are also then both positive, which makes it simpler to think about.

Answer 8

from second ratio we got
2P-Q=-2

Answer 9

or
Q=2(p+1)

Answer 10

What about using Componendo and Dividendo?

Answer 11

yes thats very useful, lol

Answer 12

yes @mathslover it works

Answer 13

so to continue on what im working
R/S=1/2 :)

Answer 14

@Marki how u got \(\dfrac{r}{s}=1/2\)

Answer 15

from second ratio
Q=2(p+1)

so
p+1/2(p+1)=R/S=1/2

Answer 16

from the proportion:
\[(p+2):(q+2)=1:2\]
I can write this one:
\[(p+2+q+2):(p+2)=-1:1\]
from which I get:
\[p+q=3p+2\]
which is compatible only with option c, sin p has to be an integer number

Answer 17

oops..since p has to be....

Answer 18

oops..error not -1:1
it is 3:1

Answer 19

@Michele_Laino (p+2+q+2):(p+2)=3:1 ?

Answer 20

yes

Answer 21

Cool.

\(\cfrac{P+2}{Q+2} = \cfrac{1}{2} \)

Using Componendo and Dividendo.

\(\cfrac{P + Q + Q + 2 }{P + 2 - Q - 2 } = \cfrac{1 + 2 }{ 1 - 2 } \)

\(\cfrac{P + 2Q + 2}{P - Q} = -3 \)

Answer 22

Oops . Mistake :

\(\cfrac{P + Q + 2 + 2 }{ P - Q } = -3 \)

Answer 23

how do i proceed

Answer 24

\( P + Q + 4 = -3(P-Q) \)
\(P + Q + 4 = -3 P + 3Q \\
-4P + 2Q = 4 \\
-2P + Q = 2\)

\(Q = 2 + 2P \)

We need to find P + Q.

So, P + Q = 2 + 2P + P = 2 + 3P

\(\cfrac{P + 1}{Q + 1} = \cfrac{R}{S} \\
\cfrac{P + 1 + Q + 1 }{ P + 1 - Q - 1 } = \cfrac{R + S}{R - S} \\
\cfrac{P + Q + 2 }{ P - Q } = \cfrac{R+S}{R-S} \)
\(\cfrac{2 + 2P + P + 2}{P - 2 - 2P} = \cfrac{R+S}{R-S}\)
\(\cfrac{3P + 4}{-P-2} = \cfrac{R+S}{R-S}\)

Hmm, still thinking.

Answer 25

i think the componenedo dividendo is tedious , it will be wise to use options

Answer 26

\( \cfrac{P + 1}{Q + 1} = \cfrac{R}{S} \\
\cfrac{P + 1 + Q + 1}{P + 1 - Q - 1} = \cfrac{R + S}{R - S} \\
\cfrac{P + Q}{P - Q} + \cfrac{2}{P-Q} = \cfrac{R+S}{R-S} \\ \)
\( \cfrac{P+Q}{P-Q} = \cfrac{R+S}{R-S} - \cfrac{2}{P-Q} \) -------- (1)

And

\(\cfrac{P+2}{Q+2} = \cfrac{1}{2} \\
\cfrac{P + Q + 4}{ P -Q } = -3 \\ \)
\( \cfrac{P+Q}{P-Q} = -3 - \cfrac{4}{P-Q} \) --------------(2)

Equate (1) and (2)

Answer 27

See if you get something? Not sure how will we use options though.

Answer 28

Let P - Q = x

\(\cfrac{R+S}{R-S} - \cfrac{2}{P -Q} = -3 - \cfrac{4}{P-Q} \)
\(\cfrac{R+S}{R-S} - \cfrac{2}{x} = -3 - \cfrac{4}{x} \)
\(\cfrac{R+S}{R-S} + 3 = \cfrac{2}{x} - \cfrac{4}{x} \)
\(\cfrac{R+S}{R-S} + 3 = \cfrac{-2}{x} \)
\(x \left( \cfrac{R+S}{R-S} + 3 \right) = -2 \)
\(x = \cfrac{-2}{ \cfrac{R+S}{ R-S} + 3} \)

Answer 29

If you do this problem in the future, it is highly recommended to do a substitution A=P+1 and B=Q+1 so that your equations look nicer: \[\Large \frac{A}{B}= \frac{R}{S} \\ \Large \frac{A+1}{B+1}=\frac{1}{2} \\ \Large solve: \ A+B-2\]

Answer 30

@Kainui still confused

Answer 31

I'm just saying do this instead so that your algebra looks cleaner, I didn't solve anything.

Answer 32

Okay, I seem to get something now.

Answer 33

by the way i was thinking of solving

\(2p-q=-2\\
p+q=3,4,5,6\)

any of those which gives integer is solution

https://www.desmos.com/calculator/owi578rpi7

Answer 34

i mean to save time

Answer 35

http://www.wolframalpha.com/input/?i=solve+%28P%2B1%29%2F%28Q%2B1%29%3DR%2FS%2C+%28P%2B2%29%2F%28Q%2B2%29%3D1%2F2%2CP%2BQ%3D5+over+integers

Answer 36

still any others methods are appriciated

Answer 37

Wow, the first equation is useless, all it tells us is that it's a ratio of two numbers since we're never given R or S. Throw it away!

Make the substitution A=P+2 and B=Q+2 and then plug them in to the second equation to get 2A=B. Remember we are trying to solve P+Q, so we plug these in and get A+B-4 that we need to solve. Use the equation I just found (2A=B) and plug it in to get A+2A-4 then we have 3A-4.

So from here we just plug in values for A since the answers are all really small, A=2 gives the smallest positive number which is 3*2-4=2. Not an answer, so try A=3 and we have 3*3-4=5. That's answer C. =)

Answer 38

lol i like it @Kainui

Answer 39

I will give medal + fan to someone who can prove to me that the first equation is not completely useless.

Answer 40

We know that there may be many solutions in integer, but if we are limited to the four choices, we can figure it as follows:
\(\dfrac{P+2}{Q+2}=\dfrac{1}{2}=\dfrac{2}{4}=\dfrac{3}{6}=...\), then
\(\dfrac{P}{Q}=\{\dfrac{-1}{0}or=\dfrac{0}{2} or=\dfrac{1}{4}=...\}\)
which means P+Q=5

Answer 41

yes that kind of thing i was looking for

Answer 42

@Kainui
I am still intrigued by the statement: " the first equation is not completely useless."
I have the impression that if P,Q,R,S \(\in\) Z, then the second equation implies the first, so the first is superfluous...
I wonder...

Answer 43

Yeah, it's sort of weird to have that first equation there, I don't see how that is at all useful. The only reason that it might be useful is that it means P is not equal to -1 maybe? But we already know from the second equation that P+2 and Q+2 are both positive from looking at the answer choices.

Answer 44

:)

Answer 45

@Kainui Thx! :)

Answer 46

\[P+Q \equiv 2 \pmod {3}\]

Answer 47

Q=2(P+1)
P+Q=3P+2 :O

Answer 48

thats the general solution for given equations

Answer 49

*diophantine equations

Answer 50

yeah so least one is 5 >.<

Answer 51

Oh yes it has to be P+Q = 3P+2