Question

help!

Answer 1

Hint: you can rewrite (x-1)^(-1) as 1/((x-1)^1) or just 1/(x-1)

Answer 2

From there, multiply every term by the inner LCD x-1 to clear out the inner fractions

Answer 3

multiply top and bottom by (x-1)

Answer 4

you got this?

Answer 5

\[\frac{3(x-1)^{-1}+7}{7(x-1)^{-1}+3}\] is your problem i think
it is really a compound fraction because \((x-1)^{-1}=\frac{1}{(x-1)}\)

Answer 6

so if you do as @juantweaver suggested, multiplying top and bottom by \((x-1)\) will clear the fractions in one step

Answer 7

\[\frac{3(x-1)^{-1}+7}{7(x-1)^{-1}+3}\times \frac{(x-1)}{(x-1)}\]
\[\frac{3+7(x-1)}{7+3(x-1)}\] then multiply out and combine like terms

Answer 8

\[2+7x-7\div7+3x-3\]?

Answer 9

i think that 2 is a typo

Answer 10

\[\frac{3+7(x-1)}{7+3(x-1)}=\frac{3+7x-7}{7+3x-3}=\frac{7x-4}{3x+4}\]

Answer 11

yes i meant 3

Answer 12

is that it?

Answer 13

yup

Answer 14

ok im going to do a similar one

Answer 15

that's what i got

Answer 16

give me a second to check

Answer 17

close, but you are off in the denominator
the numerator is \(3x+1\) but the denominator should be \(4x+1\) not \(3x+1\)

Answer 18

I think you mean 4x-1 instead of 4x+1

Answer 19

yeah that is what i meant
sorry

Answer 20

ok thank you :)