Question

Solve the following equation. Be sure to check the answer in the original equation if you multiply both sides by an expression that contains the variable.

Answer 1

\[\frac{ 3 }{ x-6 } - \frac{ 2x }{ x+1 }=\frac{ -54 }{ x^2-5x-6 }\]

Answer 2

So in this instance, I would multiply both sides by the LCD of the fractions.

Answer 3

Don't I have to factor out the last denominator?

Answer 4

Not necessarily. You can if you want to, but it's not strictly necessary. You'll see when you find the LCD.

Answer 5

Ok. I'll do it your way cuz math is too frustrating right now.

Answer 6

Have you determined the LCD?

Answer 7

uhhh, (x-6) (x+1)?

Answer 8

right. so what happens to the denominator on the right side?

Answer 9

you multiply it by x-6?

Answer 10

well
you're multiplying everything by (x-6) (x+1)

Answer 11

so since the denominator is (x-6) (x+1), it goes away

Answer 12

So, I'm left with -54

Answer 13

Do all the (x-6)(x+1)s cancel out or just the last ones?

Answer 14

I'm closing the question just to post another question while I get help with this one...

Answer 15

@sourwing , can you help us?

Answer 16

Yeah. sorry for being afk. On the right side you're left with -54. What about on the left?

Answer 17

-2x^2-15x-3?

Answer 18

hm.... check your signs...

Answer 19

your numbers are right though

Answer 20

- & + confuse me sometimes

Answer 21

lol yeah just keep it clear in your head >.>

Answer 22

I get confused sometimes too...

Answer 23

lol good cuz I'm not the only one lol

Answer 24

right so write it out
3(x+1)-(2x)(x-6)
(3x+3)-(2x^2-12)
3x+3-2x^2+12
-2x^2+3x+15

Answer 25

set that equal to -54.
Do you know how to get the answer from there?

Answer 26

yes! subtract 15 from both sides!

Answer 27

um...actually, add 54 to both sides.

Answer 28

wouldn't it be the same thing?

Answer 29

oh never mind lol

Answer 30

not necessarily. you want to set your expression equal to zero if you want to factor it and find the roots. if you add 15 to both sides, you would not find the roots of the equation.

Answer 31

2x^2+3x+69=0

Answer 32

yup. What are the roots?

Answer 33

I kind of forgot how to find those >.< lol

Answer 34

actually, I'm wrong. It's 2x^2+3x+57.

Answer 35

There's no way that this thing has rational roots. You're going to have to use the quadratic formula. >.<

Answer 36

I can use:
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
Right?

Answer 37

yeah. It's a quadratic equation, after all.

Answer 38

that can't be right ...so far, I have \[x=\frac{ -3\pm \sqrt{-447} }{ 4 }\]
Square roots cant be negative, right?

Answer 39

um...it's -2x^2. Sorry x.x

Answer 40

lol omg, ok

Answer 41

ok, so now i have \[x=\frac{ -3\pm \sqrt{465} }{ -4 }\]
Right?

Answer 42

b is 15, a is -2, and c is 57

Answer 43

try with those numbers

Answer 44

I still end up with a negative square root

Answer 45

hmm... you shouldn't. 225-(-8)(57) isn't less than one...

Answer 46

oooh ok, I got 681

Answer 47

yup. that seems all right.

Answer 48

you should get two roots. what are they?

Answer 49

I'm horrible at this part lol especially with big numbers

Answer 50

that's fine. I've figured out the numbers.

Answer 51

lol good, what are they?

Answer 52

\[x=\frac{ 15\pm \sqrt{681} }{ 4 }\]

Answer 53

It's good to be able to know the steps, but to check your answer, I'd recommend www.wolframalpha.com

Answer 54

I tried that, I don't know what the hell it was trying to give me lol

Answer 55

lol. well try those roots in the equation, and that's your work.

Answer 56

I probably put it in wrong lol

Answer 57

that site is really picky about parentheses

Answer 58

yessssss lol

Answer 59

lol
does that solve your problem?

Answer 60

I need to go soon, so if you have any questions now is a good time.

Answer 61

It says I need 2 answers

Answer 62

you do have 2 answers. it's plus/minus. one answer is 15+√681 and one is 15-√681 (the numerators that is)

Answer 63

Nevermind, I got it!!

Answer 64

all right. Bye!

Answer 65

Thank you! Bye!