|
Question 5262:
I could not find any lessons for my brother on Adding and Subtracting algebraic fractions. Could you please help me. Could you also show me how to type in fractions. In this problem x is under x+2 then you add with x+1 under x. Thank you
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! I could not find any lessons for my brother on
Adding and Subtracting algebraic fractions. Could
you please help me. Could you also show me how to
type in fractions. In this problem x is under x+2
then you add with x+1 under x. Thank you
`
x+2` ` x
--- + ---
`x ` `x+1
`
Put parentheses around the first numerator and the
second denominator because they contain 2 terms
`
(x+2)` ` `x
----- + -----
` x` ` `(x+1)
`
The LCD is x(x+1)
The first denominator is x. Thus to become equal to
the LCD it needs to be multiplied by (x+1). But we
must multiply both the numerator and the denominator
by (x+1), so that it will amount to multiplying by 1.
`
(x+2)`(x+1)` ` `x
-----·----- + -----
` x` `(x+1)` `(x+1)
`
Multiply out the (x+2)(x+1) using the FOIL method,
(x+2)(x+1) = x²+1x+2x+2 = (x²+3x+2). Do not multiply
out the denominator x(x+1), but just leave it as it is
`
(x²+3x+2) ` ` x
--------- + -----
` x(x+1)` ` (x+1)
`
The second denominator is x. Thus to become equal to
the LCD it needs to be multiplied by x. But we must
multiply both the numerator and the denominator by x,
so that it will amount to multiplying by 1.
`
(x²+3x+2)` ` `x ` `x
--------- + -----·---
` x(x+1)` ``(x+1)` x
`
Multiply the x by the x, getting x². Do not multiply
out the denominator (x+1)x, but write it as x(x+1)
`
(x²+3x+2)` ` `x²
--------- + ------
` x(x+1)` ` x(x+1)
`
Now both fractions have the same LCD, so we
combine the numerators and place it over the LCD
`
(x²+3x+2)+x²
------------
` x(x+1)
`
Remove the parentheses
`
x²+3x+2+x²
----------
` x(x+1)
Combine the x² and the x², getting 2x²
`
2x²+3x+2
--------
`x(x+1)
`
The numerator does not factor, so we are done.
`
Edwin
|
|
|
| |
