# How do you make x the subject in hxz=xh^2t?

Asked by

dotlin (

419)
November 1st, 2010

is there anyway without getting x/x or anyway so we have x as the subject?

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## 9 Answers

is 2t the exponent of h, or is it x(h^2)(t)

alright so, it’s

x times z times h = x times t times h^2

By logical inference and the properties of multiplication.

If you divide both sides by hz, you get:

x = x(something)

If that is true, then what value of (something) would that be guaranteed to be true for?

On the other hand, if (something) is not that value that would make that equation always true, then what value of x would make that true?

What you can infer from this equation, is that if the equation is true, then at least one OR the other is true, or both.

You get x=x, so x can equal anything.

@PhiNotPi How do you “get” x = x? (Not that x isn’t always x.)

@Zaku Here is how

You have the original equation h*x*z = x*h^2*t

You can divide each side by x, to get h*z=h^2*t

Since these two are equal, you can divide the left side by h*z and the right side by h^2*t

You then end up with x = x.

However, after rereading the problem, I see how the one of the x’s could be looked at as a multiplication symbol.

The problem then would be h*z=x*h^2*t

Divide each side by h^2*t

(h*z)/(h^2*t)=x

h/(h^2) * z/t = x

h/(hh) * z/t = x

1/h * z/t = x,

z/(ht) = x, which is different because of how you interpret the first x.

@PhiNotPi Ok… though taking steps to get x = x is pointless, since you already know that x = x, by definition.

Sure, there is that “the x could be multiplication” idea, too.

I was trying to make the asker think in my first answer, but I get the idea no one got what I meant.

If it’s h*x*z = x*h^2*t,

then x = x * h^2*t / z

So since we know that x = x by definition, this tells us that

h^2*t / z must not have any effect on x by being multiplied by it, therefore:

h^2*t / z = 1…

OR

x = 0.

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