Question

Help solving this rational expression..

Answer 1

\[\frac{ x+3 }{ x-3 }+\frac{ x }{ x-5 }=\frac{ x+5 }{ x-5 }\]

Answer 2

@Nnesha

Answer 3

I multiplied the first fraction by x-5 and the second by x-3

Answer 4

I added it, crossed multiply, but this doesn't like right..

Answer 5

@DanJS

Answer 6

you could do that, or you could multiply both sides by \((x-3)(x-5)\) to get rid of the fractions

Answer 7

Could you help me with my method? That's what I usually do.

Answer 8

frist multiply by each different factor top and bottom and then when the denominators are the same you can add or subtract

Answer 9

That's what i did

Answer 10

well what did you get?

Answer 11

\[\frac{ x^2-2x-15 }{ x^2-8x+15 }+\frac{ x^2-3x }{ x^2-8x+15 }=\frac{ x+5 }{ x-5 }\]

Answer 12

now you can simply combine like terms the top

Answer 13

Ok I got

Answer 14

\[\frac{ 2x^2-5x-15 }{ x^2-8x+15 }=\frac{ x+5 }{ x-5 }\]

Answer 15

right?

Answer 16

wait a sec

Answer 17

actually you were wrong for the second factor I think. How did you get x^2-3x??

Answer 18

x(x-3)

Answer 19

oh ok

Answer 20

Any more thoughts? lol

Answer 21

ya how about just like the factors both top and bottom

Answer 22

so you might be able to cross out

Answer 23

wait what was the question are we supposed to solve for x?

Answer 24

yes..

Answer 25

@satellite73

Answer 26

do think would it work if multiply both sides by (x-5) to get rid of it

Answer 27

both sides of what

Answer 28

@freckles

Answer 29

If I did this correctly i'm just stuck at

Answer 30

I dont think this is factorable

Answer 31

i believe so I don't know either lol

Answer 32

2x^2.... is not factorable with rational number

Answer 33

I dont know if it would work with replacement

Answer 34

do you have the answer choice?

Answer 35

The answer is 0,7. Just don't know how they got there.

Answer 36

x=0 and x=7 right?

Answer 37

mhm

Answer 38

give me a sec let me replace those value for x and see if its true

Answer 39

it works :(

Answer 40

I think the easiest way was to have initially aimed to clear your problem of fractions.

Answer 41

@AriPotta

Answer 42

How do I do that?

Answer 43

Can you get the solution just from my method?

Answer 44

Multiply both sides by the lcm of the bottoms
... I will have to look at your way on the desktop.

Answer 45

(x-3)(x-5), how will this remove fractions?

Answer 46

Things cancel when you divide something by itself

Answer 47

Can you draw it or use equation? Not quite seeing it

Answer 48

Oh, wait

Answer 49

Never mind

Answer 50

I know what your saying

Answer 51

yo how about subtract to make the other equal zero!!!

Answer 52

from where we left off

Answer 53

need help ??

Answer 54

@freckles

Would you say it's always a better idea to use that method instead of mine?

Answer 55

yes :p

Answer 56

\[\frac{ x+3 }{ x-3 }+\frac{ x }{ x-5 }=\frac{ x+5 }{ x-5 } \\ \text{ multiply both sides by } (x-3)(x-5) \\ (x+3)(x-5)+x(x-3)=(x+5)(x-3)\]

Answer 57

Yup, and I'm left off with x^2-7x

x(x-7)

x=7,0

:D

Answer 58

so probably it gonna be like this \[(2x^2-5x-15)/(x-5)(x-3) - (x+5)/(x-5)=0\]

Answer 59

\[(x+3)(x-5)=(x-3)(x+5)-(x-3)x \\ (x+3)(x-5)=(x-3)[x+5-x] \\ (x+3)(x-5)=(x-3)5 \\ (x+3)(x-5)=5x-15 \\ x^2-2x-15=5x-15 \\ x^2-2x-5x=0\]
yes it results into that exactly

Answer 60

now I guess you were just combining fractions?

Answer 61

x(x-7)

Answer 62

x(x-7)/(x-5)(x-3) = 0

Answer 63

Thanks everyone

Answer 64

i will post later other method that u posted above^^
bec of lag <--- take too much time to load pate

Answer 65

so you understand it now? @dtan5457

Answer 66

page ***
not pate :P

Answer 67

ok @Nnesha

Answer 68

so you got here:
\[\frac{ x+3 }{ x-3 }+\frac{ x }{ x-5 }=\frac{ x+5 }{ x-5 } \]
\[ \frac{(x+3)(x-5)}{(x-3)(x-5)}+\frac{x(x-3)}{(x-3)(x-5)}=\frac{x+5}{x-5}\]
\[\frac{x^2-2x-15+x^2-3x}{x^2-8x+15}=\frac{x+5}{x-5}\]
\[\frac{2x^2-5x-15}{x^2-8x+15}=\frac{x+5}{x-5}\]
You can do this thing that people like to call cross multiplication (really it is just multiplying stuff )
\[(2x^2-5x-15)(x-5)=(x^2-8x+15)(x+5)\]
\[2x^3-10x^2-5x^2+25x-15x+75= \\ x^3+5x^2-8x^2-40x+15x+75\]
\[x^3-15x^2+3x^2+65x-30x=0\]
\[x^3-12x^2+35x=0\]
\[x(x^2-12x+35)=0\]
\[x(x-7)(x-5)=0\]

Answer 69

This is the way I think you intended to go about it

Answer 70

you get x=0 or x=7 or x=5
but x=5 can't be a solution anyhow because it isn't in the domain of the original equation

Answer 71

How did you know 5 isn't in the domain?

Answer 72

do you know in the orginial equation you have x-5 and x-3 in the denominators right?

Answer 73

yes

Answer 74

if we have gotten x=3 or x=5 we would know either of these would be a contradiction

Answer 75

because we cannot divide by 0

Answer 76

Ohh

Answer 77

Got it

Answer 78

definitely going to use your first method though, a ton easier

Answer 79

well there is another variant of your way if you want to see that too
but it is kinda close to what we did in the method I prefer

Answer 80

so you found a common denominator for one side
you could do it for both sides though

Answer 81

find a common denominator for both sides

Answer 82

then cross out?

Answer 83

\[\frac{ x+3 }{ x-3 }+\frac{ x }{ x-5 }=\frac{ x+5 }{ x-5 } \]
like you multiply first fraction by (x-5)/(x-5)
second fraction by (x-3)/(x-3)
and we will do the third fraction by (x-3)/(x-3)

--
this was we will have the bottom on both sides
and all we have to do is set the tops equal to each other
like this:
\[\frac{x+3}{x-3} \frac{x-5}{x-5}+\frac{x}{x-5} \frac{x-3}{x-3}=\frac{x+5}{x-5} \frac{x-3}{x-3} \\ \frac{(x+3)(x-5)+x(x-3)}{(x-3)(x-5)}=\frac{(x+5)(x-3)}{(x-3)(x-5)} \\ \text{ set tops equal }\]

Answer 84

we will have the same bottoms on both side*

Answer 85

(x+3)(x-5)+x(x-3)=(x+5)(x-3)
but this is gives the exact same equation that one method gives

Answer 86

I see. Still gonna use the first method though

Answer 87

lol ok :)

Answer 88

Thanks :)

Answer 89

np :)