a/a^2-1 plus 2a/a^2-1
Equation below

Question

a/a^2-1 plus 2a/a^2-1
Equation below

bekkah323 · Answer

$$\frac{ a }{ a^2-1 } + \frac{ 2a }{ a^2-a}$$

jim_thompson5910 · Answer

the denominators are both equal to a^2-1

so you can combine the numerators and place that result over the common denominator

bekkah323 · Answer

but one is a and one is 1?

jim_thompson5910 · Answer

the numerators are _____ and _____

fill in the blanks

bekkah323 · Answer

a and 2a

jim_thompson5910 · Answer

add them

bekkah323 · Answer

3a?

jim_thompson5910 · Answer

we can do this because the denominators are the same

if they weren't the same, then we cannot add them (yet)

jim_thompson5910 · Answer

yep

jim_thompson5910 · Answer

so the final answer is 

$$\large \frac{3a}{a^2 - 1}$$

bekkah323 · Answer

i don't get why the denominators are the same because one is a^2-1 and the other is a^2-a

jim_thompson5910 · Answer

oh so that's a typo at the top then

jim_thompson5910 · Answer

you have a^2 - 1 for both at the very top

jim_thompson5910 · Answer

so the expression is 

$$\large \frac{a}{a^2 - 1} + \frac{2a}{a^2 - a}$$

right?

bekkah323 · Answer

yah i fixed it in the actual equation i wrote down, sorry?

jim_thompson5910 · Answer

that's ok

bekkah323 · Answer

yes

jim_thompson5910 · Answer

ok great

jim_thompson5910 · Answer

the first step is to factor each denominator

jim_thompson5910 · Answer

$$\large \frac{a}{a^2 - 1} + \frac{2a}{a^2 - a}$$


$$\large \frac{a}{(a-1)(a+1)} + \frac{2a}{a(a-1)}$$

jim_thompson5910 · Answer

the next step is to identify the LCD

do you see what it is?

bekkah323 · Answer

they both have an a-1 but the right side needs an a+1 and the left side needs an a?

jim_thompson5910 · Answer

very good

jim_thompson5910 · Answer

the LCD is a(a-1)(a+1)

jim_thompson5910 · Answer

the first fraction needs an 'a' in the denominator, so you have to multiply top and bottom by 'a' to get


$$\large \frac{a}{(a-1)(a+1)} + \frac{2a}{a(a-1)}$$


$$\large \frac{a*a}{a(a-1)(a+1)} + \frac{2a}{a(a-1)}$$


$$\large \frac{a^2}{a(a-1)(a+1)} + \frac{2a}{a(a-1)}$$

jim_thompson5910 · Answer

and the second fraction needs (a+1) in the denominator like you said earlier, so multiply top and bottom of the second fraction by (a+1) to get.... 

$$\large \frac{a^2}{a(a-1)(a+1)} + \frac{2a}{a(a-1)}$$ 

$$\large \frac{a^2}{a(a-1)(a+1)} + \frac{2a(a+1)}{a(a-1)(a+1)}$$ 

$$\large \frac{a^2}{a(a-1)(a+1)} + \frac{2a^2+2a}{a(a-1)(a+1)}$$

bekkah323 · Answer

ok so would the answer be$$\frac{ 3a^2 +2a}{ a(a-1)(a+1) }$$

jim_thompson5910 · Answer

correct

bekkah323 · Answer

thank you!

jim_thompson5910 · Answer

you're welcome