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i think...not sure!

I was only able to get thru math because they had tutoring at school.

A(X-Y) + X(X-Y) = 0

divide by X-Y for both term

A + X = 0

A = 0 and X =0

0(x)-0(Y) +(0^2) -0(Y) = 0

0=0

the proof works and checked out.

now that is see how it was broke down...maybe i can do it better next time..

When you get to A(x-y) + X(x-y) you actually say that it is (a+x)(x-y) foil it back out and you'll get the ax-ay+x^2-xy

Then you would get 0/(x-y) which would = 0 which would be fine, so I think reply #4 is right.

here is the correct work.

ax-ay+x^2-xy

factor out common variable a(x-y)+x(x-y)=0

divide both side by x-y =a+x=0

solve for x = x=-a

we now know what x and a is so solve for y.

a(-a)-(ay)+(-a)^2-(-ay)=0

-a^2 - ay +a^2 +ay=0

0=0

y= 0

now check our work

if x = 1

ax-ay+x^2-xy

(-1)(1) -(-1)(0) + (1)^2 -(1)(0)= 0

-1 - 0 -0 + 1 - 0 = 0

0 = 0 (check)

if x =2

ax-ay+x^2-xy

(-2)(2)-(-2*0)+(2)^2-2(0)=0

-4-0 +4-0=0

0=0 (check)

there you go. you can put any number for x and it will check out. so your answer is X= -A and Y=0.

ugh- math is so confusing!

that doesn't even make any sense - there should technically be only one way to "factor completely" to the most simplified form.

He sounds like he doesn't know what he's talking about! ;( grrr!

Math is often one of those classes you just have to try and get through.

When it says factor, It means factorise, i.e. find a common factor between different values, in this case, ax - ay + (x^2) - xy

Lets solve: ax - ay + (x^2) - xy =0

a(x-y) + x(x-y) = 0

divide both

sides by (x-y) : a + x = 0

a=0 x = 0