Question

____2a____ + _____4a_____
a + 1 a2 - 1
Add and simplify
will give medal and will fan you :)

Answer 1

is it \[\frac{ 2a }{ a+1 } + \frac{ 4a }{ a^2-1 }\]

Answer 2

?

Answer 3

yes

Answer 4

get the common factor...

Answer 5

which is 2a and 4a right?

Answer 6

no... the LCD...

Answer 7

a + 1

Answer 8

but u have a^2-1 right.. first factor that and u will get (a+1)(a-1).. so ur LCD is (a+1)(a-1)... u get me..?

Answer 9

\(\ \sf \dfrac{2a}{a + 1} + \dfrac{4a}{a^2 - 1} \)

a^2 - 1 ==> (a - 1)(a + 1)

(2a)/(a + 1) + (4a)/(a - 1)(a + 1)

Get a common denominator, multiply both sides by (a - 1)(a + 1)

\(\ \sf \dfrac{(2a) \times (a - 1)}{(a + 1)\times(a - 1)} + \dfrac{4a}{(a + 1)(a - 1)} \)

Now add them,

\(\ \dfrac{2a(a - 1)}{(a + 1)(a - 1)} + \dfrac{4a}{(a + 1)(a - 1)} \implies \dfrac{2a(a - 1) + 4a}{(a + 1)(a-1)} \implies \dfrac{a( 2(a - 1) + 4)}{(a - 1)(a + 1)}\)

Simplify 2(a - 1), 2a - 2

\(\ \sf \dfrac{a(2a - 2 + 4)}{(a - 1)(a + 1)} \implies \dfrac{a(2a + 2)}{(a - 1)(a +1)} \)

This can be further factored. 2a + 2 ==> 2(a + 1)

\(\ \sf \dfrac{a(2(a + 1))}{(a + 1)(a - 1)} \)

This cancels out, \(\ \sf \dfrac{a(2\cancel{(a + 1)}}{\cancel{(a + 1)}(a - 1)} \)
\(\ \dfrac{a(2)}{(a - 1)} \implies \) \(\ \sf \dfrac{2a}{(a - 1)} \)

Answer 10

I can see it clearly now :) both of you

Answer 11

~_^