dominion wrote:

If (y+3)(y-1) – (y-2)(y-1) = r(y-1), what is the value of y?

(1) r^2 = 25

(2) r = 5

i read a couple explanations on here that i didn't understand, like:

"(y+3)(y-1) – (y-2)(y-1) = r(y-1)

5(y-1)=r(y-1)

r=5 satisfied both conditions. " E

and

"5(y-1)=5(y-1) ==> y can be any number.

we found that: r(y-1)=5(y-1).

(1) r=+/- 5

if r=-5 -> y=1

if r=5 -> y= any number ==> we need one value only. Insuff.

(2) r=5 -> y= any number ==> we need one value only. Insuff. " E

but they don't make sense...

I'll try:

the equality itself says that r=5. does this information help us find y? no.

let's see what 1 and 2 tell us.

1. r can be +-5, but we don't care about that, do you agree?

2. r=5. once again, we only have a repetition of what is stated in the text.

neither 1 nor 2 help find y. y can be any value. let's try:

y=1 and r=5:

we have 0=0 so it's ok

let's try with 2, same thing

let's try with 3: we find always r=5.

is that a good explanation?