1. ## basic question

The expression 4/n+1 + 3/n-1 is equal to....

Thanks

2. Originally Posted by Stevo_Evo_22
The expression 4/n+1 + 3/n-1 is equal to....

Thanks
$\displaystyle \frac{4}{n+1}+\frac{3}{n-1}=?$

We want to combine these two fractions. However, their denominators are different. Multiply both top and bottom of the first fraction by n-1 and multiply both top and bottom of the second fraction by n+1:

$\displaystyle \frac{4}{n+1}\cdot{\color{red}\frac{n-1}{n-1}}+\frac{3}{n-1}\cdot{\color{red}\frac{n+1}{n+1}}$

Multiply through and simplify. Take note that $\displaystyle (n+1)(n-1)=n^2+n-n-1=\color{red}n^2-1$.

$\displaystyle \frac{4(n-1)}{n^2-1}+\frac{3(n+1)}{n^2-1}=\frac{4n-4+3n+3}{n^2-1}=\color{red}\boxed{\frac{7n-1}{n^2-1}}$

Hope this makes sense!

3. Lol actually the question was extremely easy (even for me, maths tragic). I got the same answer you did, but the problem is it's a multiple choice question and neither of them are what I had.

The options are:

A=7n-1/1-n^2
B=1-7n/1-n^2
C=7n-1/n^2-1
D=7/n^2-1
E=7/n

Is the answer that we got the same as any of these???

4. Originally Posted by Stevo_Evo_22
Lol actually the question was extremely easy (even for me, maths tragic). I got the same answer you did, but the problem is it's a multiple choice question and neither of them are what I had.

The options are:

A=7n-1/1-n^2
B=1-7n/1-n^2
C=7n-1/n^2-1
D=7/n^2-1
E=7/n

Is the answer that we got the same as any of these???
Yes.

$\displaystyle \frac{7n-1}{n^2-1}=\frac{-(1-7n)}{-(1-n^2)}=\color{red}\boxed{\frac{1-7n}{1-n^2}}$.

B would be the best answer.