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# How do you simplify (2x-5)/(x-2)+(x-8)/(x+4)

I've multiplied it out so that they both have a common denominator of (x+4)(x-2) and now I'm confused on what to do next.

### 5 Answers by Expert Tutors

Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
5.0 5.0 (3 lesson ratings) (3)
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Simplify: (2x - 5)/(x - 2) + (x - 8)/(x + 4) = f(x)

First, because you can't divide by zero,
note that x ≠ -4 or 2.

You have two fractions.
To add them you need a common denominator.

f(x) = (2x - 5)(x + 4)/((x + 4)(x - 2))
+ (x - 8)(x - 2)/((x - 2)(x + 4))

f(x) = ((2x - 5)(x + 4) + (x - 8)(x - 2))/((x - 2)(x + 4))

f(x) = (2x(x + 4) - 5(x + 4) + x(x - 2) - 8(x - 2))
/(x(x + 4) - 2(x + 4))

f(x) = (2x^2 + 8x - 5x - 20 + x^2 - 2x - 8x + 16)
/(x^2 + 4x - 2x - 8))

f(x) = (3x^2 - 7x - 4)/(x^2 + 2x - 8))

Do long division:

3x^2 - 7x - 4 | 3
3x^2 + 6x -24

-13x + 20 | 0

f(x) = 3 - (13x - 20)/((x - 2)(x + 4))

Find partial fractions:

(13x - 20)/((x - 2)(x + 4)) = A/(x - 2) + B/(x + 4)

13x - 20 = A(x + 4) + B(x - 2) = (A + B)x + 4A - 2B

A + B = 13
4A - 2B = -20
2A + 2B = 26
6A = 6
A = 1
B = 12

f(x) = 3 - (1/(x - 2) + 12/(x + 4))

f(x) = 3 - 1/(x - 2) - 12/(x + 4)

This is the simplest form of f(x) and is the one to use if you want to integrate f(x). Calculus is the goal; right?!

check:

Graph the initial form of f(x) and the last form to see if they are the same. See: http://www.wyzant.com/resources/files/265277/simplify_to_partial_fractions
Adam S. | Professional and Proficient Math TutorProfessional and Proficient Math Tutor
0
(2x-5)/(x-2)+(x-8)(x+4). The LCD is (x-2)(x+4).

Therefore to add the fractions, multiply the first fraction by (x+4)/(x+4) and multiply the second fraction by (x-2)/(x-2).

This produces: ((x+4)(2x-5) + (x-8)(x-2))/(x-2)(x+4). Multiplying and simplifying the numerator gives:
(3x^2 - 7x - 4)/(x-2)(x+4).

Factoring the numerator by using the FOIL method gives (3x-4)(x-1)/(x-2)(x+4).

Unfortunately none of the factors can be cancelled out. So we multiply the factors again to give the final expression:

(3x^2 - 7x - 4)/(x^2 + 2x -8).
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
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2X - 5   + X -8  =
X- 2       X +4

Just like Arithmetic , adding  2 unlike fractions( different denominator) :

First  Make denominator the same ,and then add:

(2X -5)(X+4)       + ( X - 8) ( X -2) =
( X -2 ) ( X +4)        ( X-2) ( X+4 )

2X^2-5X + 8X -20 +X^2-8X -2X+16
( X -2 ) ( X +4)

3X^2-7X -4
X^2 +2X -8

Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Chani;
[(2x-5)/(x-2)]+[(x-8)/(x+4)]
Let's take the first part of the expression, and multiply the numerator and denominator by (x+4)/(x+4)
[(2x-5)/(x-2)][((x+4)/(x+4)]
Let's FOIL the numerator...
(2x-5)(x+4)
FIRST...(2x)(x)=2x2
OUTER...(2x)(4)=8x
INNER...(-5)(x)=-5x
LAST...(-5)(4)=-20
2x2+3x-20
Let's take the second part of the expression, and multiply the numerator and denominator by (x-2)/(x-2)
[(x-8)/(x+4)][(x-2)(x-2)]
(x-8)(x-2)
Let's FOIL the numerator...
FIRST...(x)(x)=x2
OUTER...(x)(-2)=-2x
INNER...(-8)(x)=-8x
LAST...(-8)(-2)=16
x2-10x+16
We know that the denominators of both bracketed expressions is the same, [(x-2)(x+4)].  Therefore, we can add the numerators together...
2x2+3x-20
x2-10x+16
3x2-7x-4

(3x2-7x-4)/[(x-2)(x+4)]

Samantha G. | SAT Prep and Psychology TutorSAT Prep and Psychology Tutor
0
You're multiplying the addend by ((x+4)/(x+4)) and the second addend by ((x-2)/(x-2)). You've got that part right! So now both addends have a common denominator: (x+4)(x-2).

So now you can add the numerators together: ((2x-5)(x+4))+((x-8)(x-2)) and that's all over (x-2)(x+4). Now you're going to forget all about the denominator and focus on that numerator.

You need to FOIL both (2x-5)(x+4) and (x-8)(x-2) and then add everything together. Once you've simplified that, put it over (x-2)(x+4) and you'll have your answer.

(Reminder: FOIL = first, outside, inside, last)