Math Is Fun Forum

Just a wild guess, but would you sub the values of x and y into the first equation?

Thanks, I got;

x=2a-11/(a-3)

I got
x-z+ 6/(a-3) =2

I got y=3/a-3

I'm not sure but is it;

2x+5y-ax-2ay+2a=3

ay-3y=3

Edit: oh ok, I see that I wasn't supposed to expand it.

So, would it be wrong if I removed x and solved for z in the first equation and got

z=2y+2 ?

Okay that cleared some stuff up, so would I start off by rearranging the three equations?

Sorry, I'm really terrible at these questions.

Hello, no it was right.

Thanks for helping gAr; post #1 was the same question.

Yep, that's right.
The second should be 5/14

If you're referring to the above question, then no that's there was. However, there is a separate part of that question which asks.

"If one bowl is chosen at random and from it one piece of fruit is chosen at random without replacement. Find the probability that the fruit chosen is an apple."
and that..
"...if the piece of fruit chosen was an apple. Find the probability that A was the chosen bowl".

That's weird, my text book shows a different answer....