Can this fraction be simplified further

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Discussion Overview

The discussion revolves around the simplification of the expression $\frac{2a}{a^2-4} - \frac{1}{a-2}$. Participants explore the process of combining fractions, identifying the least common denominator (LCD), and simplifying the resulting expression. The focus includes mathematical reasoning and technical explanations related to algebraic manipulation.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant identifies the difference of squares in $a^2-4$ as $(a+2)(a-2)$ and asks if the fraction can be simplified further.
  • Another participant questions the least common denominator (LCD) needed to combine the fractions.
  • Several participants provide steps to combine the fractions, indicating the need to make denominators equal.
  • There is a correction regarding an error in subtraction during the simplification process, leading to a new expression $\frac{a-2}{(a+2)(a-2)}$.
  • Participants discuss the cancellation of terms in the expression, with one participant questioning how $a+2$ moved from the denominator to the numerator.
  • A later reply acknowledges a mistake in the cancellation process and clarifies that the final expression is $\frac{1}{(a+2)}$, while also noting the restrictions on the variable $a$.

Areas of Agreement / Disagreement

Participants generally agree on the process of finding the LCD and combining the fractions, but there is disagreement regarding the correctness of the simplification steps and the implications of the restrictions on the variable $a$. The discussion remains unresolved in terms of a definitive conclusion about the simplification process.

Contextual Notes

Participants mention that $a$ cannot be equal to -2 or 2, which affects the validity of the final expression. There is an acknowledgment that the expression $\frac{1}{(a+2)}$ implies a restriction on $a$ that needs to be considered.

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

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 :)
 
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What is your LCD if you are going to combine the two fractions as indicated by subtraction?
 
MarkFL said:
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 ? :)
 
mathlearn said:
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
 
MarkFL said:
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 ? :)
 
mathlearn said:
$$\frac{2a}{(a+2)(a-2)} - \frac{a+2}{(a+2)(a-2)}=\frac{2a-(a+2)}{(a+2)(a-2)}=\frac{2a-a-2}{(a+2)(a-2)}=\frac{a-2}{(a+2)(a-2)}$$

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

Correct ? :)

How did $a+2$ move from the denominator to the numerator?
 
MarkFL said:
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)
 
mathlearn said:
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
 

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