# rearrange equation

• May 18th 2008, 06:45 AM
Turple
rearrange equation
How would you rearrange:

v=1/3x²h to make x the subject?

Thanks
• May 18th 2008, 06:46 AM
Moo
Hello,

Is it $v=\frac 13 \cdot x^2h$ or $v=\frac{1}{3x^2h}$ ?
• May 18th 2008, 06:51 AM
Turple
It's a third :)

Also found another 2:

1. It's just the quadratic formula without the +- thing to make c the subject.
2. S=2пr²+2пrh to make h the subject

:)

Thanks
• May 18th 2008, 10:08 AM
Simplicity
$v=\frac 13 \cdot x^2h$

Multiply both sides by $3$:

$3v = x^2 h$

Divide both sides by $h$:

$\frac{3v}{h} = x^2$

Square root both side:

EDIT: $\pm \sqrt{\frac{3v}{h}} = x$
• May 18th 2008, 10:09 AM
Moo
Quote:

Originally Posted by Air
[...]

Square root both side:

${\color{red}\pm} \sqrt{\frac{3v}{h}} = x$

Be careful :p
• May 18th 2008, 10:20 AM
Simplicity
Quote:

Originally Posted by Turple
It's a third :)

Also found another 2:

1. It's just the quadratic formula without the +- thing to make c the subject.
2. S=2пr²+2пrh to make h the subject

:)

Thanks

1.)

$x = \frac{-b \sqrt{b^2 - 4ac}}{2a}$

$2ax = -b \sqrt{b^2 - 4ac}$

$-\frac{2ax}{b} = \sqrt{b^2 - 4ac}$

$\left(-\frac{2ax}{b}\right)^2 = b^2 - 4ac$

$\left(-\frac{2ax}{b}\right)^2 - b^2 = -4ac$

$\frac{(-\frac{2ax}{b})^2 - b^2}{-4a} = c$

$c = \frac{b^2 - (\frac{2ax}{b})^2}{4a}$

2.)

$S=2 \pi$ $r^2+2 \pi rh$

$S - 2 \pi r^2 = 2 \pi rh$

$\frac{S - 2 \pi r^2}{2 \pi r} = h$
• May 18th 2008, 10:23 AM
Simplicity
Quote:

Originally Posted by Moo
Be careful :p

Edited. (Blush)
• May 18th 2008, 01:56 PM
collating terms
"How would you rearrange:
v = 1/3 x² h to make x the subject?"

Having seen some of the replies, I wonder if I am missing something here. Do you mean how to I arrange to solve for x?

Assuming you meant (1/3)x and not 1/(3x), first multiply every term by 3 to eliminate fractions:
3v = 3(1/3)x² h
So:
3v = x2 h
Or since you want to isolate x:
x² h = 3v
Divide by coefficient of x:
(x² h)/h = 3v/h
Giving:
x² = 3v/h
Transform by taking square root of both sides:
x = +sqrt(3v/h) and –sqrt(3v/h)

The 1/(3x) would have been more interesting but just as easy.