Question

i am doing something wrong and was hoping someone could correct me
solve,simplify and leave in factored form

2(x-1)^-1 + 3 / 3(x-1)^-1 +2

Answer 1

what steps have you taken?

Answer 2

i made the first parts fraction that look like 1 3
------ + -------
2(x-1) 1

Answer 3

thats a start :)

1
------- + 3
2(x-1)
------------
1
------ + 2
3(x-1)

\[\frac{\frac{1}{2(x-1)}+3}{\frac{1}{3(x-1)}+2}\]

Answer 4

we can multiply this by a useful form of "1"; lets try (x-1)/(x-1) to get:

\[\frac{\frac{1}{2}+3(x-1)}{\frac{1}{3}+2(x-1)}\]

Answer 5

now to clear the fractions I would multiply by 2/2 and 3/3 to get

\[\frac{\frac{1}{2}+3(x-1)}{\frac{1}{3}+2(x-1)}*\frac{2}{2}=\frac{1+6(x-1)}{\frac{2}{3}+4(x-1)}\]

\[\frac{1+6(x-1)}{\frac{2}{3}+4(x-1)}*\frac{3}{3}=\frac{3+18(x-1)}{2+12(x-1)}\]
we can factor top and bottom to get:

\[\frac{3(1+6(x-1))}{2(1+6(x-1))}\]
and cancel like terms to end up with: 3/2 if i did it right

Answer 6

i see something I might have done wrong ..

Answer 7

i got 3/2 but the book got 3x-1/2x+1

Answer 8

2(x-1)^-1 .... notice that the exponent only applies to the (...) part; so leave the 2 up and drop the (x-1) down ...

Answer 9

lets revise our work now:

\[\frac{\frac{2}{(x-1)} + 3}{\frac{3}{(x-1)}+2}*\frac{(x-1)}{(x-1)}=\frac{2+3(x-1)}{3+2(x-1)}\]

Answer 10

we might want to expand the work to:

\[\frac{2+3x-3}{3+2x-2}\]
\[\frac{3x-1}{2x+1}\]

Answer 11

yep, that was our mistake :)

Answer 12

i don't get the part where you wrote"we can multiply this by a useful form of "1"; lets try (x-1)/(x-1) to get:"

Answer 13

one of the most useful tricks in algebra is to add zero; or multiply by 1.

the values change how an equation LOOKS while keeping the VALUE of the equation the same..

Answer 14

still trying to understand, i feel dumb

Answer 15

we could also think of this as equating the expression to some unknown value; say "N" for a number of "S" for a solution and algebra our way thru it as well like this:

\(S = \cfrac{\cfrac{2}{(x-1)} + 3}{\cfrac{3}{(x-1)} +2}\); lets multiply both sides by \(\cfrac{3}{(x-1)}+2\)


\(S\left(\cfrac{3}{(x-1)}+2\right)=\left(\frac{2}{(x-1)}+3\right)\); expand it out to ....


\(\cfrac{S3}{(x-1)}+S2=\frac{2}{(x-1)}+3\); multiply thru by (x-1) to clear the fractions


\(S3+S2(x-1)=2+3(x-1)\)

lets factor the S back out of the left side there:

\(S(3+2(x-1))=2+3(x-1)\); divide out (3+2(x-1)) to solve for "S"

\(S=\cfrac{2+3(x-1)}{3+2(x-1)}\); expand it

\(S=\cfrac{2+3x-3}{3+2x-2}\); and simplify

\(S=\cfrac{3x-1}{2x+1}\)

Answer 16

algebra is all about being able to move things around without changing any inherent values

Answer 17

think of algebra like shuffling a deck of cards; the number of cards never changes, but you are able to put them in any order by shuffling them.

algebra allows us to shuffle thru an expression or equation so the we can see it in a new way that might be easier to deal with than the original statements. It retains the same value; just changes its form