Question

For all real numbers x and y, if x # y = x(x-y), then x # (x # y)=
This is from the Princeton Review book for the GRE. Please help!

Answer 1

it looks like they defined an operator
and want you to use that definition
x # y = x(x-y)

it looks that that says, take the left operand (the x) times (left - right)

so in
x # (x#y) you take the left operand: x times (left - right) or
x * (x - (x#y))
we can now replace x#y with x*(x-y) to get
x * (x - (x*(x-y)))

that can be "simplified" or re-written a number of ways... if this is multiple choice you may have to tweak it to match one of the answers.

Answer 2

I'm still having trouble understanding your explanation. What do you mean by (left - right). And what does # signify in the question?

Answer 3

the question "made up" the operator #

they are saying: if you see a # b then you can replace that with a*(a-b)
where a is what I mean by the left operand (because a is on the left side of #) and b is the right operand.

Answer 4

It is a "rule"

if you see left # right then replace it with
left * (left - right)

for example

Answer 5

So everytime I see "x", I need to replace it with "x(x-y)" ?

Answer 6

? no, everywhere you see x#y you replace with x(x-y)

Answer 7

when y is "complicated" example y is (x#y)
then
x# (x#y) becomes x * (x - (x#y) )

Answer 8

which can be further reduced by applying the rule again
x# (x#y) becomes x * (x - (x#y) )

x * (x - (x#y) ) becomes
x * (x - x(x-y))

Answer 9

So this is what I understand so far:
Replace "x#y" with "x(x-y)"
So x # (x#y) becomes:
x# [x(x-y)].
Now how do I get rid of x# ?

Answer 10

the same way. to keep things clear let A stand for x(x-y)
so you have
x#A
which becomes x(x-A)
now replace A with x(x-y) to get
x(x-x(x-y))

Answer 11

My final question then is if you're able to substitute "x(x- )" for "x#" , why can you disregard the "#y" portion of x#y? My natural instinct was to solve for x, which would have moved #y to the other side of the equation, which is obviously wrong, but I don't understand how you can infer that x# = x(x-