Finding Y in Terms of X - Step by Step

[0000]

¿How to Find Y?

Hi, I have this equation and I need to find Y in terms of X. Could anyone explain me, step by step, how to do it?

X=(Y-2)/3 - 1/3Y

Sorry if there is any grammar mistakes, english isn't my native language.

 

[Repetit]

Y cancels out as you see:

X = (Y-2)/3 - 1/3 Y = Y/3 - 2/3 - Y/3 = -2/3

 

[mathman]

X=(Y-2)/3 - 1/3Y

There is a little ambiguity, but I'll assume that the last term is1/(3Y), not (1/3)Y (repetit assumption). In that case, multiply all terms by Y and you will get a quadratic equation in Y, with X as part of the coefficient of the linear term. Solve the quadratic for Y and you will have your answer (actually two answers).

 

[0000]

Thanks, it was 1/(3Y). My fault.

 

[berkeman]

Welcome to the PF, 0000. Please be sure to post homework and coursework questions like this one in the Homework Help forums here on the PF (where I've moved this thread to), and not in the general forums.

Now, can you write out the quadratic equation as mathman has suggested, and show us how you would solve it?

 

[0000]

ok

Ok, berkeman

X=(Y-2)/3 - 1/(3Y)

yx=y((y-2)/3) - 1/(3Y))

yx=(y^2 - 2y)/3 -y/3y)

yx=(y^2 - 2y)/3 - 1/3

yx=(y^2 - 2y - 1)/3

y^2 - 2y - 1 - 3yx=0

y^2 - y(2+3x) - 1 = 0

a=1
b=-(2+3x) = (-2 - 3x)
c=-1

y=((2+3x)±√(9x^2 + 12x + 8))/2

 

[CompuChip]

yx=(y^2 - 2y)/3 - 1/3
yx=(y^2 - 2y - 1)

You forgot a 3 there. It's correct on the next line though.

y^2 - 2y - 1 - 3yx=0
y^2 - y(2-3x) - 1 = 0

Are you sure? Check what happens if you work out the brackets in the second line. I think it's a minus sign off.

The idea is correct though, just watch the minuses :)