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Solve for $h$:

$$\frac{h+z}{h} = \frac{a}{b}$$

I'm not sure how to simplify this with cross multiplication.

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closed as off-topic by KReiser, José Carlos Santos, Namaste, Key Flex, Deepesh Meena Sep 28 '18 at 1:38

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  • $\begingroup$ Is the left side $h + \frac{z}{h}$ or $\frac{h+z}{h}$? $\endgroup$ – Sean Roberson Sep 27 '18 at 21:33
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    $\begingroup$ $\frac {h+z}{h} = \frac ab$ or $h + \frac {z}{h} = \frac ab$? Formatting is important. $\endgroup$ – Doug M Sep 27 '18 at 21:36
  • $\begingroup$ (h+z)/h = a/b sorry about that $\endgroup$ – Katelyn Sep 27 '18 at 21:42
  • $\begingroup$ Using MathJax could avoid confusion like this. $\endgroup$ – KM101 Sep 27 '18 at 22:29
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$$\frac{h+z}{h} = \frac{a}{b}$$

$$\implies b(h+z) = ah$$

$$bh+bz = ah \implies bz = ah-bh \implies bz = (a-b)h \implies h = \frac{bz}{a-b} $$

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Multiply both sides of the equation by $h$ and you get: $\\h^2-\frac{a}{b}h+z=0$ This should be easy to solve as it's just a quadratic.

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$$\frac{h^2}{h}+\frac{z}{h} = \frac{a}{b}$$ $$\frac{h^2+z}{h} = \frac{a}{b}$$ $$h^2+z = \frac{a}{b}h$$ $$h^2-\frac{a}{b}h+z=0$$ Using the quadratic formula, we can get $$h = \frac{\frac{a}{b}\pm\sqrt{\left(\frac{a}{b}\right)^2-4z}}{2}$$

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