Question

What did I do wrong while trying to solve this complex fraction? (I'll show all my work)

Answer 1

k

Answer 2

i will help him @Nnesha

Answer 3

\[\frac{ \frac{ 1 }{ z-7 }+\frac{ 7 }{ z } }{ \frac{ z }{ z^2-7z }-1 }\]

Answer 4

I multiplied the numerator by x(x-7)

Answer 5

and got \[\frac{ 8z-49 }{z^2-7z } \]

Answer 6

Then for the bottom part i multiplied the -1 by the same thing : x(x-7)

Answer 7

I got\[\frac{ z }{ z^2-7z }-\frac{ -z^2+7z }{ z^2-7z }=\frac{ -z^2-6z }{ z^2-7z }\]

Answer 8

hey wait do you mean multiply by z(z-7)

Answer 9

I put these on top of each other and flipped the 2nd term to get

\[\frac{ 8z-49 }{ z^2-7z }\times \frac{ z^2-7z }{ -z^2-6z }\]

Answer 10

and yes

Answer 11

now i'm stuck with \[\frac{ 8z-49 }{ -z^2-6z }\]

Answer 12

which isn't the right answer

Answer 13

based on my work, what did i do wrong?

Answer 14

someone please enlighten me ive been stuck on this for a few hours

Answer 15

\[\frac{ \frac{ 1 }{ z-7 }+\frac{ 7 }{ z } }{ \frac{ z }{ z^2-7z }-1 }\times \frac{z(z-7)}{z(z-7)}\]
\[=\frac{z+7(z-7)}{z-z(z-7)}\]

Answer 16

\[\frac{ \frac{ 1 }{ z-7 }+\frac{ 7 }{ z } }{ \frac{ z }{ z^2-7z }-1 }\]\[= \dfrac{\frac{z + 7 (z - 7)}{z(z-7)}}{\frac{z - z(z - 7)}{z(z - 7)}} \]\[= \dfrac{z + 7(z - 7)}{z - z(z -7)}\]\[= \dfrac{z + 7z - 49}{z - z^2 + 7z}\]\[= \dfrac{8z - 49}{ 8z-z^2}\]

Answer 17

remove the parentheses in the numerator and the denominator get
\[\frac{z+7z-49}{z-z^2+7}\] cleans up to
\[\frac{8z-49}{-z^2+z+7}\]

Answer 18

i multiplied the numerator part only the 2nd term by z(z-7) and the first term by z

Answer 19

so they have the same denominator?

Answer 20

oh i made a mistake
@ParthKohli has it better then me

Answer 21

\[\frac{8z-49}{-z^2+z+7}\] is wrong it is
\[\frac{8z-49}{-z^2+z+7z}\]

Answer 22

well i got the numerator correct

Answer 23

but the bottom part i have -z^2-6z

Answer 24

before the cleaning up the denominator is what @ParthKohli wrote
\[z-z(z-7)\]

Answer 25

I don't know is this correct for the denominator?

\[\frac{ z }{ z^2-7z }-1=\frac{ -z^2+7z }{ z^2-7z }\]

Answer 26

I put -1 as -1/1 then multiplied that with z^2-7z

Answer 27

so now

\[\frac{ z }{ z^2-7z }-\frac{ -z^2+7z }{ z^2-7z }=\frac{ -z^2-6z }{ z^2-7z }??\]

Answer 28

@Jhannybean

Answer 29

@saifoo.khan

Answer 30

@DanJS

Answer 31

@AlexandervonHumboldt2

Answer 32

\[\frac{ z }{ z^2-7z }-\frac{ -z^2+7z }{ z^2-7z }=\frac{ -z^2-6z }{ z^2-7z }\]

you are doing .... \[\frac{ z }{ z^2-7z }-(-1)\]

instead of doing: \[\frac{ z }{ z^2-7z }-1\]

you are taking the negative twice which makes it wrong

Answer 33

if i multiply -1 by z^2-7z won't that change the signs?

Answer 34

\[\frac{ z }{ z^2-7z }-1\\=\frac{ z }{ z^2-7z }+(-1)\\=\frac{ z }{ z^2-7z }+{(-1)(z^2-7z)\over z^2-7z}\\=\frac{ z }{ z^2-7z }\color{red}+{\color{red}-z^2\color{red}+7z \over z^2-7z}\]
\[=\large{\color{red}-z^2\color{red}{+8}z \over z^2-7z}\]

Answer 35

o.

Answer 36

so if i can take the new numerator and denominator i get

\[\frac{ 8z-49 }{ -z^2+8z }\]

Answer 37

\[\checkmark\]

Answer 38

on my regents book the numerator is 7z-47..

Answer 39

the denominator is correct

Answer 40

They can get that if they subtracted on the numerator but I don't see that being possible?

Answer 41

maybe you are looking at a different question ...to get numerator 7z-47 you have to have \[{2\over z(z-7)} + \frac7z\] ....for the numerator

Answer 42

There's is definitely a 1 instead of a 2...

Answer 43

Book error???

Answer 44

i guess..

Answer 45

that's pretty crazy....most absurd question..seriously.

Answer 46

Thanks for clearing up my mistake on the denominator though.

Answer 47

np