Question

Solve and check for extraneous solutions.

(5x+20)/(x^2+3x) - (x-2)/(x+3) = 6/x

Answer 1

that is fun...

Answer 2

??????

Answer 3

ok so first. what do you need to add or subtract fractions?

Answer 4

a common denominator

Answer 5

ok so (this is only one way of going about this) what is the LCM of (x+3) and \((x^2+3x)\)

Answer 6

x+3

Answer 7

ok so now, can you tell me how you got that?

Answer 8

factored x out of x^2+3x

Answer 9

ok so now, what do you have to multiply (5x+20)/(x^2+3x) by to get a x+3 on the bottom, or what do you have to multiply (x-2)/(x+3) by to get a x^2+3x on the bottom?

Answer 10

x?

Answer 11

or 1/x

Answer 12

ok I presonally prefer to make the common denominator x^2+3x here because who wants to deal with another fraction right?

Answer 13

but so, you have to multiply by 1, you can't just multiply by x, so what do you have to multiply by?

Answer 14

1/x? I'm confused

Answer 15

ok, if you had 2/3+3/4 what would you do?

Answer 16

multiply by 12

Answer 17

well, you wouldn't multiply by 12

Answer 18

you would have to get a common denominator of twelve, yes,

Answer 19

but how would you add those, can you show me your work for that?

Answer 20

Sorry I had to step away for a minute
2/3 + 3/4
1/4(2/3) + 1/3(3/4)
2/12 + 3/12 = 5/12

Answer 21

it's ok, so there is an error in your work there, can you read this for me then tel me what it is? http://www.purplemath.com/modules/fraction4.htm

Answer 22

4/4(2/3) + 3/3(3/4)

Answer 23

8/12 + 9/12 = 17/12

Answer 24

right so now, what does 4/4 simplify to?

Answer 25

1

Answer 26

so do I multiply my problem by x/x?

Answer 27

yes!

Answer 28

progress! lol ok I'll work on that for a minute

Answer 29

ok, just remember though it is only one fraction being multipled

Answer 30

wait I messed up, I'll try again
(5x+20)/(x^2+3x) - (x^2-2x)/(x^2+3x) = 6/x

Answer 31

that's right

Answer 32

ok so now, make the left one big old fraction

Answer 33

(-x^2+3x+20)/(x^2+3x) = 6/x

Answer 34

I messed up again :(
(-x^2+7x+20)/(x^2+3x) = 6/x ??

Answer 35

messing up is ok, let me look

Answer 36

your second is correct :)

Answer 37

I just get the feeling, are you a nontraditional student?

Answer 38

yay!

Answer 39

I do school online if that's what you mean

Answer 40

oh ok, I thought you may be older returning to college type

Answer 41

Oh no I'm only 17 haha

Answer 42

lol ok, so anyways after here. you can get rid of the fraction on the right

Answer 43

by multiplying with x/x again?

Answer 44

or does it need the same denominator as the left side?

Answer 45

not quite you just need to multiply by x to cancel out the x on the right

Answer 46

(-x^2+7x+20)/(x^2+3x) = 6

Answer 47

don't I still need to get rid of the fraction on the left? do I multiply everything by the denominator?

Answer 48

not yet, first you have to make sure you multiply both sides by x

Answer 49

i already multiplied the right side by x though

Answer 50

sorry, had to go. but you are undoing the fraction 6/x, in order to do that you have to multiply both sides by x not just one

Answer 51

that's fine!
so then:
x[(-x^2+7x+20)/(x^2+3x)]=(6/x)(x)

Answer 52

yes

Answer 53

(-x^3+7x^2+20x)/(x^2+3x) = 6

Answer 54

now do I multiply both sides by x^2+3x?

Answer 55

no, now you need to reduce the fraction

Answer 56

I think I worked it out to the answer... x = 2 or x = -1

Answer 57

and x = 0 was an extraneous solution

Answer 58

x can't equal zero, what does your def of extraneous mena?

Answer 59

mean?

Answer 60

I have to go take a final but here is the rest, sorry. \[\frac{x(-x^2+7x+20)}{x(x+3)} = 6\]\[\frac{(-x^2+7x+20)}{(x+3)} = 6\]

Answer 61

factor the top and some cancelling should happen