Please help me with algebra?

Question

[ ( 3 / (x-2)) - ( 1 / (x-3)) ] / [ ( 1 / (2-x)) + (2 / (x+3)) ]

I hope it makes sense with the brackets..it looks a little messy. Please help. thanks :)

Answer 1

It's nice to see someone actually putting in all the brackets for a change - it may look awkward, but it means we don't have to guess at where the divisions start and end.

[(3 / (x-2)) - (1 / (x-3))] / [(1 / (2-x)) + (2 / (x+3))]
First step is to put them into common denominators and simplify:
= [(3 (x-3) - 1 (x-2)) / ((x-2) (x-3))] / [(1 (x+3) + 2 (2-x)) / ((2-x) (x+3))]
= [(2x - 7) / ((x-2) (x-3))] / [(7 - x) / ((2-x) (x+3))]
Now convert from division to multiplication by turning the second fraction upside down:
= [(2x - 7) / ((x-2) (x-3))] × [((2-x) (x+3)) / (7 - x)]
Now x-2 = -(2-x), so we can cancel them out; we'll get rid of the negative sign by putting it into the 7-x factor to make it x-7:
= [(2x-7) / (x-3)] × [(x+3) / (x-7)]
= [(2x-7) (x+3)] / [(x-3) (x-7)].

There are no common factors, so that's the simplest form; if you like you can multiply out the factors to get
(2x^2 - x - 21) / (x^2 - 10x + 21).
This can also be written as
2 + (19x - 63) / (x^2 - 10x + 21).

Answer 2

just working with the numerator:

3/(x-2) - 1/(x-3), multiply the left term by (x-3)/(x-3) and the right term by (x-2)/(x-2) to get

(3x - 9 - x +2)/[(x-3)(x-2)] = (2x -7)/[(x-3)(x-2)]

Now just working with the denominator:

1/(2-x) + 2/(x+3), notice that (2-x) = -(x-2), so make that substitution

-1/(x-2) + 2/(x+3), now multiply the left term by (x+3) and the right term by (x-2) to get:

[-(x+3) + 2(x-2)]/[(x-2)(x+3) = (x-7)/[(x-2)(x+3)]

We then multiply the new fraction with the inverse of the denominator, ie [(x-2)(x+3)]/(x-7) to get:

[(2x-7)(x-2)(x+3)]/[(x-3)(x-2)(x-7)]

we can cancel out the (x-2)/(x-2) to get:

[(2x-7)(x+3)]/[(x-3)(x-7)]

errors are possible, please check the work

Answer 3

[(3/(x-2)}-1/(x-3)}]/ [(1/(2-x)}+(2/(x+3)}]
3/(x-2) -1/(x-3)
={3(x-3)-1(x-2)}/(x-2)(x-3)
=(3x-9-x+2)/(x-2)(x-3)
=(2x-7)/(x-2)(x-3)
1/(2-x) +2/(x+3)
= -1/(x-2)+2/(x+3)
={-1(x+3)+2(x-2)}/(x-2)(x+3)
=(-x-3+2x-4)/(x-2)(x+3)
=(x-7)/(x-2)(x+3)
Therefore the given expression
=(2x-7)/(x-2)(x-3) divided by (x-7)/(x-2)(x+3)
=(2x-7)/(x-2)(x-3)*(x-2)(x+3)/(x-7)
=(2x-7)(x+3)/(x-7)

Answer 4

x/20

Answer 5

Yes, is little messy - I try but I so confuse - sorry