Forget the name of it, but i need a review

  • Thread starter austin1250
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Sorryx2-1In summary, the conversation is about solving a given equation by finding a common denominator and adding the fractions. The process involves multiplying the numerators and denominators by appropriate factors to get the common denominator and then adding the fractions. There may be confusion over the correct method of finding the common denominator, but ultimately the solution is found by adding the fractions with the common denominator.
  • #1
austin1250
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Homework Statement



I basiclly just need a little review and maybe some 1 to show me how to solve out the equation/and equation like it. here is an example:

2x+1 / x-1 + 1/ x+1 = ?

Homework Equations


?



The Attempt at a Solution


?
 
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  • #2
austin1250 said:

Homework Statement



I basiclly just need a little review and maybe some 1 to show me how to solve out the equation/and equation like it. here is an example:

2x+1 / x-1 + 1/ x+1 = ?
I assume by ? you mean a number like 3 or 2 or even some function f(x)?but here's a clue

[tex]\frac{1}{a} + \frac{1}{b}= \frac{a+b}{ab}[/tex]
 
  • #3
by Question mark i mean it would equal like 2x + 1 / x^2 - 1 or something along those lines
 
  • #4
austin1250 said:
by Question mark i mean it would equal like 2x + 1 / x^2 - 1 or something along those lines



then just bring them to the same common denominator (x-1)(x+1). Do you know how to do that?
 
  • #5
I THINK you mean you want to add the two fractions (2x+1)/(x-1)+ 1/(x+1).
(Please use parentheses!)

As Rock.freak667 said, you need to get "common denominators". Here, the common denominator is (x-1)(x+1). Multiply numerator and denominator of the first fraction by x+1 and numerator and denominator of the second fraction by x-1.
 
  • #6
rock.freak667 said:
then just bring them to the same common denominator (x-1)(x+1). Do you know how to do that?

no i forget, do you foil it ?
 
  • #7
HallsofIvy said:
I THINK you mean you want to add the two fractions (2x+1)/(x-1)+ 1/(x+1).
(Please use parentheses!)

As Rock.freak667 said, you need to get "common denominators". Here, the common denominator is (x-1)(x+1). Multiply numerator and denominator of the first fraction by x+1 and numerator and denominator of the second fraction by x-1.

yeah, ok I did that and i got

(2x^2-x-1)/(x2-2x+1) + (x+1)/ (x^2 -2x-2)

now what
 
  • #8
austin1250 said:
yeah, ok I did that and i got

(2x^2-x-1)/(x2-2x+1) + (x+1)/ (x^2 -2x-2)

now what

the denominator would be (x-1)(x+1)

now do as HallsofIvy said and

HallsofIvy said:
Multiply numerator and denominator of the first fraction by x+1 and numerator and denominator of the second fraction by x-1.
 
  • #9
rock.freak667 said:
the denominator would be (x-1)(x+1)

now do as HallsofIvy said and

that just confused me. i did what halls said and i got the answer you quoted. so i did it wrong?
 
  • #10
austin1250 said:
that just confused me. i did what halls said and i got the answer you quoted. so i did it wrong?

[tex]\frac{2x+1}{x-1} + \frac{1}{x-1}[/tex]



[tex]= \frac{(2x+1)}{(x-1)(x+1)} + \frac{(x+1)}{(x-1)(x+1)}[/tex]

since the denominators are both the same, you can just add the numerators now.
 
  • #11
rock.freak667 said:
[tex]\frac{2x+1}{x-1} + \frac{1}{x-1}[/tex]



[tex]= \frac{(2x+1)}{(x-1)(x+1)} + \frac{(x+1)}{(x-1)(x+1)}[/tex]

since the denominators are both the same, you can just add the numerators now.

did you copy that down wrong? its

(2x+1)/(x-1)+ 1/(x+1).
 
  • #12
austin1250 said:
did you copy that down wrong? its

(2x+1)/(x-1)+ 1/(x+1).

Sorry there, forgot to do some more typing, should be this



[tex]
= \frac{(2x+1)(x+1)}{(x-1)(x+1)} + \frac{(x-1)}{(x-1)(x+1)}
[/tex]
 
  • #13
Now, austin1250, use "FOIL" to multiply (x-1)(x+1) carefully! It is NOT [itex]x^2- 2x- 1[/itex].
 
  • #14
HallsofIvy said:
Now, austin1250, use "FOIL" to multiply (x-1)(x+1) carefully! It is NOT [itex]x^2- 2x- 1[/itex].

oh did i say that? Sorry

x2-1
 

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