Question

Simplify. The result should involve exactly 1 ratio.

Answer 1

(1/x - x/ x+1) / (2/x+1 - x-1/x)

Answer 2

\[\frac{ \frac{ a }{ b } }{ \frac{ c }{ d } }=\frac{ a }{ b }\times \frac{ d }{ c }\]

Answer 3

can you please help?

Answer 4

simplify it? because I am confused...

Answer 5

Combine the fractions in the top of the ratio first...
\[\frac{ 1 }{ x }-\frac{ x }{ x+1 }\]
The LCD is x(x+1) so
\[\frac{ 1 }{ x }\left( \frac{ x+1 }{ x+1 } \right)-\frac{ x }{ x+1 }\left( \frac{ x }{ x} \right)\]
Now multiply those out and simplify .... then we'll go from there.

Answer 6

x+1/x / xˆ2+x - xˆ2 / xˆ2 + x. is this correct?

Answer 7

Yes but I would leave the denominator unmultiplied for later. use...\[\frac{ x+1 }{ x \left( x+1 \right)}-\frac{ x^2 }{ x \left( x+1 \right) }\]
but now combine the two fractions since they have common den.

Answer 8

What do you get?

Answer 9

x+1 - xˆ2 / x(x+1) right?

Answer 10

Well yes ... I thin you mean (x+1-x^2)/[x(x+1)]. I know this seems picky but the order of operations are a pretty picky on this...it is equivalent to:\[\frac{ x+1-x^2 }{ x \left( x+1 \right) }\]

Answer 11

Now for sake of convention I would suggest writing the terms of the numerator in descending order of power...so that would be:\[\frac{ -x^2+x+1 }{ x \left( x+1 \right) }\]

Answer 12

yes, I understand this

Answer 13

Now follow the same procedure for the fractions in the bottom of the ratio (keeping the denominator factored instead of multiplied out)

Answer 14

What do you get?

Answer 15

2xˆ2 + 2 - xˆ2 -1 / x(x+1). Is this correct?

Answer 16

Please not that this is written very poorly.

(1/x - x/ x+1) / (2/x+1 - x-1/x)

I believe you mean:

(1/x - x/ (x+1)) / (2/(x+1) - (x-1)/x)

That's a whole different ballpark.

Answer 17

@Paynesdad

Answer 18

I agree is the bottom of the ratio:\[\frac{ 2 }{ x+1 }-\frac{ x-1 }{ x }\]

Answer 19

It is not quite what you wrote at the beginning but it makes the most sense so I am assuming you forgot some parentheses.

Answer 20

@Paynesdad how do I continue for the down ratio? can you please help?

Answer 21

Is the bottom ratio the one I listed a couple of lines above? (I can't tell from your post as it seems to be missing some key parentheses.)

Answer 22

yes, that is the bottom ratio

Answer 23

Ok so what is the LCD of the bottom of the ratio then?

Answer 24

x+1?

Answer 25

Not quite x+1 is part of it but not the whole thing. Look at what you did for the top of the ratio.

Answer 26

x(x+1) ?

Answer 27

There you go so what do you need to multiply the top and bottom of the first fraction by so that it has the lcd?

Answer 28

x/x?

Answer 29

for the first fraction

Answer 30

yep and the second?

Answer 31

x+1?

Answer 32

well (x+1)/(x+1) yes...so go ahead and multiply those out (not the denominators though) and combine the fractions and tell me what you get.

Answer 33

2x/xˆ2+x - xˆ2+x/xˆ2 +2. is it correct?

Answer 34

No the first one is good.

Answer 35

Now you have\[-\frac{ \left( x-1 \right)\left( x+1 \right) }{ x \left( x+1 \right)}\] ... also don't worry about multiplying out the denominator ...it is best to just leave it "unmultiplied" (or in factored form). What does the numerator multiply to?

Answer 36

this is the second part of the second fraction right? because I am confused

Answer 37

Yes.

Answer 38

Ok I understand now. it the numerator 2x - (x-1)(x+1)?

Answer 39

yep there you go

Answer 40

so multiply that and simplify. what do you get?

Answer 41

i dont know:(

Answer 42

@Paynesdad

Answer 43

Well multiply (x-1)(x+1)... can you do that?

Answer 44

xˆ2-1

Answer 45

@Paynesdad can you please help?

Answer 46

yes so now you can combine the numerators of the two fractions since their denominators are the same... right?

Answer 47

what do you mean by combine? add them or ...?@Paynesdad

Answer 48

@Paynesdad

Answer 49

yes add, subtract them do what every the expression indicates to do.

Answer 50

-xˆ2+x-1 + 2x - xˆ2 -1 / x(x+1). is this correct?

Answer 51

whoa too much....let's back up...you had...

Answer 52

\[\frac{ 2x }{ x \left( x+1 \right) }-\frac{ x^2-1 }{ x \left( x+1 \right) }\] after multiplying out each fraction right?

Answer 53

yes

Answer 54

ok so when you combine those fractions you subtract the numerators and the denominator stays the same...just like fractions with numbers...so what is that

Answer 55

2x-xˆ2-1/ x(x+1)

Answer 56

good now again I would suggest putting the numerator's terms in descending power order. So

Answer 57

Well wait one thing ... it was 2x - (x^2-1), right?

Answer 58

yes

Answer 59

Well this why those parentheses are so important...2x-(x^2-1) does not equal 2x - x^2 -1 does it?

Answer 60

no, it equals 2xˆ4 -2 right?

Answer 61

No that s way off... remember to distribute the minus sign on the stuff in parentheses!...With \[\frac{ 2x }{ x \left( x+1 \right) }-\frac{ x^2-1 }{ x \left( x+1 \right) }\] you are not just subtracting the x^2 but x^2 and the -1! so\[\frac{ 2x }{ x \left( x+1 \right) }-\frac{ x^2-1 }{ x \left( x+1 \right) }=\frac{ 2x-\left( x^2-1 \right) }{ x \left( x+1 \right) }\] so \[\frac{ 2x }{ x \left( x+1 \right) }-\frac{ x^2-1 }{ x \left( x+1 \right) }\neq\frac{ 2x- x^2-1 }{ x \left( x+1 \right) }\] do you see what I am getting at?

Answer 62

@RH ?

Answer 63

is the answer (x-1)ˆ2 / (x+2)ˆ2

Answer 64

@Paynesdad

Answer 65

No...you seem to have gone off the tracks...let's go back to\[\frac{ 2x }{ x\left( x+1 \right) }-\frac{ x^2 -1}{ x \left( x+1 \right) }=\frac{ 2x- \left( x^2-1 \right)}{x \left( x+1 \right) }\] when you distribute the "-" sign what do you get?

Answer 66

-2xˆ4 + 2x? @Paynesdad

Answer 67

How are you getting \[-2x^4\] ???

Answer 68

-2xˆ3 sorry. is it correct?

Answer 69

@Paynesdad

Answer 70

No because you are not multiplying! That is subtraction so you distribute the minus sign to get\[\frac{ 2x -\left( x^2-1 \right)}{ x \left( x+1 \right)}=\frac{ 2x-x^2+1 }{ x \left( x+1 \right) }\]

Answer 71

Not sure why you were multiplying.

Answer 72

ok now I understand, sorry for my mistake. how do i continue to simplify?

Answer 73

:-) ok I would put the numerator in power order just for the sake of convention...so that would be\[\frac{ -x^2 +2x+1}{ x \left( x+1 \right) }\] so now lets put these back into the ratio...So we have...

Answer 74

-xˆ2 + x - 1/ x(x+1) divided by -xˆ2 + 2x -1/ x(x+1)

Answer 75

Yes! so now reciprocate the fraction in the bottom of the ratio and cancel what can be and what are you left with?

Answer 76

I am not sure I understand what you mean with that? :(

Answer 77

Ok first of all I think you have a couple of -1 that should be +1 double check that

Answer 78

where???? @Paynesdad

Answer 79

So you should have \[\frac{ \frac{ -x^2+x+1 }{ x \left( x+1 \right) } }{ \frac{ -x^2+2x+1 }{ x \left( x+1 \right) } }\] ... look closely at everything and make sure you agree

Answer 80

@RH

Answer 81

yes!!!! you are right!!! the +1! sorry

Answer 82

Ok so you can reciprocate the bottom of the ratio and get...

Answer 83

\[\left( \frac{ -x^2+x+1 }{ x \left( x+1 \right) } \right)\left( \frac{ x \left( x+1 \right) }{ -x^2+2x+1 } \right)\]

Answer 84

do i cancel out now the same x(x+1)?

Answer 85

Yep!

Answer 86

and how do I simplify after that?

Answer 87

Unless you can factor the polynomials in the numerator and denominator (which you can't) and find common factors to cancel. You are done.

Answer 88

I would suggest learning the proper place to put parentheses when using / instead of a fraction bar. ... i.e. \[\frac{ 3+2 }{ 4+1 }=(3+2)/(4+1)\] not 3+2/4+1 because \[3+2/4+1=3+\frac{ 1 }{ 2 }+1=4\frac{ 1 }{ 2 }\]

Answer 89

thanks a lot!!!