1. ## need an explanation

hi, i need to know sometheing guys

i have an equation here, which reads: x = 2/(y-3)^2 - 1
which then, according to the answer goes to: x = 2/y^2 - 6y + 8

can anyone tell me how the 6y and 8 in the second equation come about, because i don't understand where they come from??

thank you,
ryry

2. Originally Posted by diablo
hi, i need to know sometheing guys

i have an equation here, which reads: x = 2/(y-3)^2 - 1
which then, according to the answer goes to: x = 2/y^2 - 6y + 8

can anyone tell me how the 6y and 8 in the second equation come about, because i don't understand where they come from??

thank you,
ryry
x = 2/[(y-3)^2 - 1]
because that's the only thing that makes sense with what you are saying.

Look at just the denominator:
(y - 3)^2 - 1

I'm going to multiply out the (y - 3)^2:
(y - 3)^2 = (y - 3)(y - 3) = y(y - 3) - 3(y - 3) = y^2 - 3y - 3y + 9 = y^2 - 6y + 9

So
(y - 3)^2 - 1 = y^2 - 6y + 9 - 1 = y^2 - 6y + 8

-Dan

3. Originally Posted by diablo
hi, i need to know sometheing guys

i have an equation here, which reads: x = 2/(y-3)^2 - 1
which then, according to the answer goes to: x = 2/y^2 - 6y + 8

can anyone tell me how the 6y and 8 in the second equation come about, because i don't understand where they come from??

thank you,
ryry
When you expand (y-3)^2 - 1 you get( y^2-6y+9) - 1 = y^2 - 6y + 8.

This can also calculate as (y-3)^2 - 1^2 = (y-3-1)(y-3+1) = (y-4)(y-2) = y^2-2y-4y+8 = y^2 - 6y + 8

4. wow, thats real good stuff guys,

i understand now, and i have a couple more equations if you're not too bothered. i'm not super when it comes to algebra, but i'm learning...

1. its the same equation, but earlier, there is a transition from √2/x+1 = y - 3 to 2/x +1 = (y-3)^2. i understand squaring y-3 but i thought you had to square both sides to remove the square root.

2. again, what process do you go through in your head, when going from 2/x = (y-3)^2 to x = 2/(y-3)^2

3. and the process going from B = x/2∂ x R to B x R = x/2∂

thanks a lot people. your help is appreciated.

5. Next time make a new thread.
Originally Posted by diablo
1. its the same equation, but earlier, there is a transition from √2/x+1 = y - 3 to 2/x +1 = (y-3)^2. i understand squaring y-3 but i thought you had to square both sides to remove the square root.
sqrt(2/(x+1))=y-3
Then,
2/(x+1)=(y-3)^2
Correct.
sqrt(2/x)+1=y-3
Then you cannot square.
Subtract 1 from both sides,
sqrt(2/x)=y-4
Then attempt to square,
2/x=(y-4)^2
2. again, what process do you go through in your head, when going from 2/x = (y-3)^2 to x = 2/(y-3)^2
Reciprocal of both sides.
Then multiply by two to remove the 2 from denominator.

3. and the process going from B = x/2∂ x R to B x R = x/2∂
Multiply both sides by R, then the R's cancel on the right hand side.

6. thank you, that was lots of help. is there any websites you know of which practice Qs and As for algebra???