# Having sign problem with algebraic fractions

• Jan 26th 2013, 06:48 PM
KevinShaughnessy
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
• Jan 26th 2013, 08:53 PM
chiro
Re: Having sign problem with algebraic fractions
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)]?
• Jan 26th 2013, 09:37 PM
Soroban
Re: Having sign problem with algebraic fractions
Hello, Kevin!

They approached the problem differently . . . that's all.

Quote:

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

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

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

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

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

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

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

. . $=\;\frac{13-5v}{6(v-1)(v+1)}$
• Jan 26th 2013, 11:17 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
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)].
• Jan 26th 2013, 11:19 PM
KevinShaughnessy
Re: Having sign problem with algebraic fractions
Quote:

Originally Posted by Soroban
Hello, Kevin!

They approached the problem differently . . . that's all.

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

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

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

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

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

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

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

But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive?
• Jan 27th 2013, 01:18 AM
earthboy
Re: Having sign problem with algebraic fractions
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?

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