Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having trouble solving this expression: $$\frac{(x - 1)(7x + 6)}{(x - 1)(x + 1)^2 }-\frac{ 7}{ (x + 1)}$$

What's the steps to solve this?

I know you expand $(x + 1)^2$ to $(x + 1)(x + 1)$,

and that you need to find a common denominator before adding the numerators together.

The final answer is $\quad\displaystyle\frac{-1}{x^2 + 2x + 1}.$

Thanks.

share|improve this question
    
I think someone got the problem wrong: looks like it should be a $-$, not a $+$. –  Ron Gordon Feb 7 '13 at 4:39
add comment

2 Answers

up vote 4 down vote accepted

We want to simplify the following:

$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)}$$

First, we cancel the common factor $(x-1)$ in the left-hand fraction:

$$\frac{(x-1)(7x + 6)}{(x-1)(x+1)^2} - \frac {7}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7}{(x+1)}$$

Now we find a common denominator to add the fractions, and see that we want both denominators to be $(x+1)^2$. To accomplish this, we can multiply the numerator and the denominator of the second fraction by the factor of $(x+1)$:

$$\frac{(7x + 6)}{(x+1)^2} - \frac{7}{(x+1)} \cdot \frac{(x+1)}{(x+1)} = \frac{(7x + 6)}{(x+1)^2} - \frac {7(x+1)}{(x+1)^2}$$ $$ = \frac{(7x + 6 - 7(x+1))}{(x+1)^2} \;=\;\frac{ -1}{(x+1)^2} $$ $$\;=\;\frac{-1}{x^2 + 2x + 1}$$

share|improve this answer
    
How did you get (x+1)^2 on the right hand side? –  Cypras Feb 7 '13 at 4:46
    
Because both denominators: $(x+1)$ and $(x+1)^2$ - divide $(x+1)^2$. Notice that $\dfrac{7(x+1)}{(x+1)^2} = \dfrac{7}{(x+1)}$ –  amWhy Feb 7 '13 at 4:47
    
Ohh okay. Sort of starting to understand. –  Cypras Feb 7 '13 at 4:50
    
Why not by (x+1)^2? –  Cypras Feb 7 '13 at 4:53
    
Yeah I see you times by (x+1) to the right hand fraction, but shouldn't it be (x+1)^2, you times 7 by (x+1) not (x+1)^2, I don't understand why. –  Cypras Feb 7 '13 at 5:00
show 6 more comments

If x$\neq$1, you can cancel $(x-1)$ in the numerator and denominator of $\frac{(x - 1)(7x + 6)}{(x - 1)(x + 1)^2 }$ so, $$\frac{(x - 1)(7x + 6)}{(x - 1)(x + 1)^2 }-\frac{ 7}{ (x + 1)}=\frac{7x + 6}{(x + 1)^2 }-\frac{ 7}{ (x + 1)}$$

Now, if x$\neq$-1 you can multiply and divide $\frac{7}{x + 1}$ by $(x+1)$ and add the fractions:

$$\frac{7x + 6}{(x + 1)^2 }-\frac{ 7}{ (x + 1)}=\frac{7x + 6}{(x + 1)^2 }-\frac{7(x+1)}{ (x + 1)^2}=\frac{7x + 6-7(x+1)}{(x + 1)^2}=\frac{-1}{(x + 1)^2}$$

share|improve this answer
    
Why do you not times the right fraction by (x+1)^2? –  Cypras Feb 7 '13 at 4:57
    
@Cypras, because i need have a denominator in the two fractions such that both are equal, this, because you know that if i have two fractions $\frac{A}{B}$ and $\frac{C}{B}$ then, the sum of the 2 fractions is $\frac{A+C}{B}$ and so it is easier to add, (it is equivalent to searching the common denominator, which is the same thing amWhy said) –  dwarandae Feb 7 '13 at 5:06
    
@amWhy, sorry, I started writing before, but when I published the answer I had not yet seen another answer. –  dwarandae Feb 7 '13 at 5:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.