# Solving for z

• May 8th 2012, 07:40 AM
Builderjay2011
Solving for z
g=h(x+y/z-x) solve for z

The solution I keep getting is...

z= hx+hy-gx/-g

should this be the correct answer?
• May 8th 2012, 07:46 AM
princeps
Re: Solving for z
If you mean :

$\displaystyle g=h\cdot\left(\frac{x+y}{z-x}\right)$

then solution should be :

$\displaystyle z=x+\frac{h(x+y)}{g}$
• May 8th 2012, 07:49 AM
Builderjay2011
Re: Solving for z
The book show something totally different
• May 8th 2012, 08:03 AM
Prove It
Re: Solving for z
Quote:

Originally Posted by Builderjay2011
The book show something totally different

1. What does the book show?

2. What have you tried?
• May 8th 2012, 08:20 AM
Builderjay2011
Re: Solving for z
How do you all show the formating of your formulas? Makes it easier for me to show my work to you.
• May 8th 2012, 09:10 AM
Prove It
Re: Solving for z
It's called LaTeX. There is a LaTeX subforum on this site where you can get some help on the coding.
• May 8th 2012, 09:37 AM
Builderjay2011
Re: Solving for z
Ok here's what I did....

g=h(x+y/x-z) for z

multiply both sides by (x-z)

g(x-z) = h(x+y)

gx-gz-gx = hx+hy-gx (clear parentheses and subtract "gx" from both sides)

-gz = hx+hy-gx

z = hx+hy-gx/-g

Let me know if this makes sense.
• May 8th 2012, 06:17 PM
Prove It
Re: Solving for z
Quote:

Originally Posted by Builderjay2011
Ok here's what I did....

g=h(x+y/x-z) for z

multiply both sides by (x-z)

g(x-z) = h(x+y)

gx-gz-gx = hx+hy-gx (clear parentheses and subtract "gx" from both sides)

-gz = hx+hy-gx

z = (hx+hy-gx)/-g

Let me know if this makes sense.

With the necessary modification, what you have is fine.