Can this fraction be simplified further

[mathlearn]

$\frac{2a}{a^2-4} - \frac{1}{a-2}$

Now using difference of squares in $a^2-4$ = $\left(a+2\right) \left(a-2\right)$

Then the fraction would be $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{1}{a-2}$

Can this fraction be simplified further?

Many Thanks :)

 

[MarkFL]

What is your LCD if you are going to combine the two fractions as indicated by subtraction?

 

[mathlearn]

What is your LCD if you are going to combine the two fractions as indicated by subtraction?

That would be to make denominators equal

Then the fraction would be $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{1*\left(a+2\right)}{a-2*\left(a+2\right)}$

Which now equals in $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{\left(a+2\right)}{(a-2)\left(a+2\right)}$

Which I guess would result in $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{\left(a+2\right)}{(a-2)\left(a+2\right)}$

$ \frac{a+2}{(a-2)\left(a+2\right)}$

Correct ? :)

 

[MarkFL]

That would be to make denominators equal

Then the fraction would be $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{1*\left(a+2\right)}{a-2*\left(a+2\right)}$

Which now equals in $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{\left(a+2\right)}{(a-2)\left(a+2\right)}$

Which I guess would result in $\frac{2a}{\left(a+2\right) \left(a-2\right)} - \frac{\left(a+2\right)}{(a-2)\left(a+2\right)}$

$ \frac{a+2}{(a-2)\left(a+2\right)}$

Correct ? :)

Yes, you have the correct LCD:

$$\frac{2a}{(a+2)(a-2)} - \frac{1}{a-2}\cdot\frac{a+2}{a+2}=\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}$$

But, you made an error in your subtraction:

$$\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$$

Now, all that's left is to divide out any factors common to the numerator and denominator. :D

 

[mathlearn]

Yes, you have the correct LCD:

$$\frac{2a}{(a+2)(a-2)} - \frac{1}{a-2}\cdot\frac{a+2}{a+2}=\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}$$

But, you made an error in your subtraction:

$$\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$$

Now, all that's left is to divide out any factors common to the numerator and denominator. :D

$$\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$$

$\frac{\cancel{a-2} }{(a+2) \cancel {(a-2)} } = (a+2)$

Correct ? :)

 

[MarkFL]

$$\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$$

$\frac{\cancel{a-2} }{(a+2) \cancel {(a-2)} } = (a+2)$

Correct ? :)

How did $a+2$ move from the denominator to the numerator?

 

[mathlearn]

How did $a+2$ move from the denominator to the numerator?

Oh! That's a mistake :D

$\displaystyle \frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$

$\frac{\cancel{a-2} }{(a+2) \cancel {(a-2)} } =\frac{1}{(a+2)} $

Many Thanks MarkFL (Smile) (Party)

 

[topsquark]

Oh! That's a mistake :D

$\displaystyle \frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$

$\frac{\cancel{a-2} }{(a+2) \cancel {(a-2)} } =\frac{1}{(a+2)} $

Many Thanks MarkFL (Smile) (Party)

You almost have it...We have a slight problem. The expression 1/(a + 2) comes out all right, but the original expression says that a can't be either -2 or 2. We have to list that in the final answer as well. So your solution is 1/(a + 2), a not equal to -2, 2.

(Technically 1/(a + 2) means we can't have a = -2 anyway, so all we really need is a not equal to 2.)

-Dan