Forget the name of it, but i need a review

[austin1250]

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

?

 

[rock.freak667]

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

\frac{1}{a} + \frac{1}{b}= \frac{a+b}{ab}

 

[austin1250]

by Question mark i mean it would equal like 2x + 1 / x^2 - 1 or something along those lines

 

[rock.freak667]

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?

 

[HallsofIvy]

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.

 

[austin1250]

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 ?

 

[austin1250]

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

 

[rock.freak667]

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

Multiply numerator and denominator of the first fraction by x+1 and numerator and denominator of the second fraction by x-1.

 

[austin1250]

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?

 

[rock.freak667]

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

\frac{2x+1}{x-1} + \frac{1}{x-1}



= \frac{(2x+1)}{(x-1)(x+1)} + \frac{(x+1)}{(x-1)(x+1)}

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

 

[austin1250]

\frac{2x+1}{x-1} + \frac{1}{x-1}



= \frac{(2x+1)}{(x-1)(x+1)} + \frac{(x+1)}{(x-1)(x+1)}

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).

 

[rock.freak667]

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



<br /> = \frac{(2x+1)(x+1)}{(x-1)(x+1)} + \frac{(x-1)}{(x-1)(x+1)}<br />

 

[HallsofIvy]

Now, austin1250, use "FOIL" to multiply (x-1)(x+1) carefully! It is NOT x^2- 2x- 1.

 

[austin1250]

Now, austin1250, use "FOIL" to multiply (x-1)(x+1) carefully! It is NOT x^2- 2x- 1.

oh did i say that? Sorry

x²-1