Question

(x^2-xy)/(y-x). I know the answer is -x I just don't understand why.

Answer 1

did you factor x from the top ?

and factor -1 from the bottom

Answer 2

\[\frac{x^2-xy}{y-x}\]
\[\frac{x^2-xy+y^2-y^2-xy+xy}{y-x}\]
\[\frac{(x-y)(x-y)-y^2+xy}{y-x}\]
\[\frac{(y-x)(x-y)+y^2-xy}{y-x}\]
\[(x-y)+\frac{y^2-xy}{y-x}\]
\[(x-y)+\frac{y(y-x)}{y-x}\]
\[(x-y)+y\]
review that for mistakes of course

Answer 3

phis idea might be simpler ... but not as spectacular ;)

Answer 4

spectacular is not the adjective that springs to mind...

Answer 5

but it does work...

Answer 6

but there is a minus sign missing...

Answer 7

in the 4th step i pulled it out of the square and for some reason thought it distrbuted over the -y^2 + xy ... in error

Answer 8

@amistre64 Although spectacular looking,
I still do not understand your explanation... :c and @phi i did what you said but I don't see how it gets me -x. D:

Answer 9

jayycee post your work

Answer 10

my method amounts to completing a square is all; which you may not have covered yet

Answer 11

(x(x-xy))/(-1(-y+x))

Answer 12

if we distribute x(x-xy) with get x^2 -x^2 y which not what you started with

redo the top

Answer 13

x(x-y) / -1 (-y+x) ?

Answer 14

yes, now x(x-y) = x^2 - xy which matches.

notice you can write -y+x as x+ -y or x-y

can you simplify what you have...

Answer 15

Sooo, x(x-y) / _y +x ? but then what?

Answer 16

?

you start with
\[ \frac{x(x-y)}{-1(-y+x) } \]
or , writing -y+x as x-y
\[ \frac{x(x-y)}{-1(x-y) } \]

remember anything divided by itself is 1.

Answer 17

OH! how did I forget!. Thank you so much!

Answer 18

do you know how to multiply fractions? top times top and bottom times bottom

that means that
\[ \frac{x}{-1} \cdot \frac{(x-y)}{(x-y)} = \frac{x(x-y)}{-1(x-y)} \]
but what is
\[ \frac{(x-y)}{(x-y)} \] ?

Answer 19

just one

Answer 20

so you are left with x/-1 which is -x
(one way is multiply top and bottom by -1 to get -x/1 or just -x)

Answer 21

is something unclear ?

Answer 22

Nope! i completely understand!:D