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Where does the Y go ???
- Thread starter maris
- Start date
J
JeffM
Guest
I think you need to give us more information. The question as posed is not intelligible to me.
maris said:4x-2xy+1/(x-1)
I know how to work this out but where does the Y go ??
maris,
did you mean to type (4x - 2xy + 1)/(x - 1), as in
\(\displaystyle \frac{4x - 2xy + 1}{x - 1} \ \ ?\)
lookagain said:maris said:4x-2xy+1/(x-1)
I know how to work this out but where does the Y go ??
maris,
did you mean to type (4x - 2xy + 1)/(x - 1), as in
\(\displaystyle \frac{4x - 2xy + 1}{x - 1} \ \ ?\)
YES!
J
JeffM
Guest
Marismaris said:lookagain said:maris said:4x-2xy+1/(x-1)
I know how to work this out but where does the Y go ??
maris,
did you mean to type (4x - 2xy + 1)/(x - 1), as in
\(\displaystyle \frac{4x - 2xy + 1}{x - 1} \ \ ?\)
YES!
The expression already looks simplified to me. The y is there: it can't just go away. I keep thinking that something else is in the problem statement. Or perhaps if it just says simplify, it's a trick question because it already is simplified.
(4x-1) (x-1) = 4x^2 - 5x + 1; sure don't look like 4x - 2xy + 1 :shock:maris said:The problem says it comes out to (4x-1) (x-1)/ (x-1) .....
Dennis, when you factor it comes out to that?? Im totally confused now.Denis said:(4x-1) (x-1) = 4x^2 - 5x + 1; sure don't look like 4x - 2xy + 1 :shock:maris said:The problem says it comes out to (4x-1) (x-1)/ (x-1) .....