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Find an equation of variation in which $y$ varies jointly as $t$ and $r$ and inversely as the square of $w$, and $y=2$ when $t=4, r=9,$ and $w=6$.

I can set this up as two separate equations, $y=ktr$, and $y=\dfrac k{w^2}$. Then I can:

  • solve it as $k=\dfrac 1{18}$ for the first equation and $k=72$ for the second equation.
  • Set the $y's $ equal to each other, which gives me $ktr=\dfrac k{w^2}$, which simplifies to $trw^2=1$, which is incorrect.

Both seem wrong; how do I solve this?

1 Answers1

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I would combine them together into a single equation, with a single constant $K$ to solve for: $$ y = K \cdot \frac{tr}{w^2} $$

Adriano
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