Having sign problem with algebraic fractions

Hi,

I'm having a problem with an algebraic fractions equation. It goes:

5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1)

The first thing I do is factor out the negative in the first fraction, getting:

-5/3(1-v)

Giving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors:

-5*2(v+1)

6(3v-1)

3(1-v)

Which gives me -10v -10 + 18v -6 -3v + 3

Which adds up to 5v - 13

BUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction.

Thanks,

Kevin

Hey KevinShaughnessy.

Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?

Hello, Kevin!

There is nothing wrong with your work or your answer.
They approached the problem differently . . . that's all.
Both answers are correct.

Quote:

---

$\displaystyle \text{Simplify: }\:\frac{5}{3(v-1)} + \frac{3v-1}{(1-v)(1+v)} + \frac{1}{2(v+1)}$

---

They factored a "minus" out of the second fraction.

. . $\displaystyle \frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)} $


The LCD is $\displaystyle 6(v-1)(v+1)\!:$

. . $\displaystyle \frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}} $

. . $\displaystyle =\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)} $

. . $\displaystyle =\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$

. . $\displaystyle =\;\frac{13-5v}{6(v-1)(v+1)}$

Quote:

---

Originally Posted by chiro

Hey KevinShaughnessy.

Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]?

---

It's (3v-1) / [(1+v)(1-v)].

Quote:

---

Originally Posted by Soroban

Hello, Kevin!

There is nothing wrong with your work or your answer.
They approached the problem differently . . . that's all.
Both answers are correct.


They factored a "minus" out of the second fraction.

. . $\displaystyle \frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)} $


The LCD is $\displaystyle 6(v-1)(v+1)\!:$

. . $\displaystyle \frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}} $

. . $\displaystyle =\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)} $

. . $\displaystyle =\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$

. . $\displaystyle =\;\frac{13-5v}{6(v-1)(v+1)}$

---

But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?

Quote:

---

Originally Posted by KevinShaughnessy

But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?

---

perhaps you misread the answer and you should check it again...
your answer was$\displaystyle \frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\displaystyle \frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $\displaystyle (1-v)$ to $\displaystyle (v-1$) by multiplying the numerator and denominator by $\displaystyle -1$.
(Happy)

Quote:

---

Originally Posted by earthboy

perhaps you misread the answer and you should check it again...
your answer was$\displaystyle \frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\displaystyle \frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $\displaystyle (1-v)$ to $\displaystyle (v-1$) by multiplying the numerator and denominator by $\displaystyle -1$.
(Happy)

---

It has all become clear hah. Thank you everyone!