# need an explanation

• Oct 4th 2006, 09:21 AM
diablo
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
• Oct 4th 2006, 09:36 AM
topsquark
Quote:

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
• Oct 4th 2006, 09:39 AM
OReilly
Quote:

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
• Oct 4th 2006, 10:03 AM
diablo
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.
• Oct 4th 2006, 10:09 AM
ThePerfectHacker
Next time make a new thread.
Quote:

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
Quote:

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.

Quote:

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.
• Oct 4th 2006, 12:35 PM
diablo
thank you, that was lots of help. is there any websites you know of which practice Qs and As for algebra???