adding and subtracting rational expressions?

Question

x ~ 5x
__ + ___
x+5 ~ x-5

could anyone solve this and explain how they did it to me? i always get stuck after finding a common denomenator

the swirlyish line is just there to keep the numbers separated; they're separating the two equations

Answer 1

.. x ..... 5x
----- + ------ =
x+5 .... x-5

... x(x-5) ....... 5x(x+5)
------------- + ------------ =
(x+5)(x-5) .. (x-5)(x+5)

x² - 5x + 5x² + 25x
------------------------ =
.... (x+5)(x-5)

6x² + 20x
--------------
(x+5)(x-5)

Answer 2

In this problem, the two denominators have no common factors so the least common denominator is the product:
(x+5)(x-5). Multiply top and bottom of each fraction by the missing factor:
[x(x-5)]/[(x+5)(x-5)] + [5x(x+5)]/[(x-5)(x+5)] =

[x(x-5) + 5x(x+5)]/[(x+5)(x-5)] =

[x^2 - 5x +5x^2 +25x]/[(x+5)(x-5)] =

(6x^2 +20x)/[(x+5)(x-5)]


In general to find the least common demoninator, factor each denominator completely, write down each distinct factor and raise it to the highest power it occurs in any one denominator.

Example with numbers: LCD of 24 and 36:

48 = (2^4)*3 36 = (2^2)*(3^2)

Distinct factors: 2 and 3, highest power of 2 is 4, highest power of 3 is 2. So the LCD is (2^4)*(3^2) = 16*9 = 144

Answer 3

(x / (x + 5)) + (5x / (x - 5))

Multiply left by (x - 5) / (x - 5) and right by (x + 5) / (x + 5):

[(x(x - 5)) / ((x + 5)(x - 5))] + [(5x(x + 5)) / ((x - 5)(x + 5))]

Common denominator is (x + 5)(x - 5). let's look at the numerators:

x(x - 5) + 5x(x + 5)
= x^2 - 5x + 5x^2 + 25x
= 6x^2 + 20x
= 2x(3x + 10)

Final answer:

2x(3x + 10) / ((x + 5)(x - 5))

Answer 4

im assuming you mean something like x 5x
---- + -------
x+5 x-5

the best way is to multiply by the denominators

(x-5)x (x+5)(5x)
------------- + ------------
(x+5)(x-5) (x+5)(X-5)

then just multiply the tops, add the fractions, simplify, and woohoo, you're done

(6x^2 + 20x)
-----------------
(X-5)(X+5)