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I've been banging my head against the wall for a while now:

$x = s^2 - t^2$

$y = s + t$

$z = s^2 + 3t$

Express $z$ in terms of $x$ and $y$.

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Hint: What is $\dfrac{x}{y}$ equal to?

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  • $\begingroup$ $x/y$ = $s - t$ I've done that and tried incorporating it but just don't see how it works. $\endgroup$ – Sir Winford Apr 27 '11 at 3:44
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    $\begingroup$ If you know $s-t$ and $s+t$ then surely you can find $s$ and $t$. $\endgroup$ – Emre Apr 27 '11 at 3:50
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    $\begingroup$ Follow up question - what's $(s + t) + (s - t)$ equal to? $\endgroup$ – Hans Parshall Apr 27 '11 at 3:59
  • $\begingroup$ @Hans: great question (+1) $\endgroup$ – Fabian Apr 27 '11 at 5:35
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$$x=s^{2}-t^{2}=(s+t)(s-t)$$ so $$s+t=\frac{x}{s-t}$$ $$s-t=\frac{x}{s+t}=\frac{x}{y}$$

$$(s+t) + (s-t) = 2s=\frac{x}{s-t}+\frac{x}{y}$$ $y=s+t$, so $t=y-s$ and therefore: $$2s=\frac{x}{2s-y}+\frac{x}{y}=x(\frac{1}{2s-y}+\frac{1}{y})$$ $$2s=x(\frac{y}{2sy-y^2}+\frac{2s-y}{2sy-y^2})=x(\frac{2s}{2sy-y^2})$$ $$1=\frac{x}{2sy-y^2}$$ $$2sy-y^2=x$$ $$2sy=x+y^2$$ $$s=\frac{x+y^2}{2y}$$ From $z=s^2+3t$ we have: $$z=(\frac{x+y^2}{2y})^{2}+3t$$ $y=s+t$ so $t=y-\frac{x+y^2}{2y}$ and finally: $$z=(\frac{x+y^2}{2y})^{2}+3(y-\frac{x+y^2}{2y})$$ Pretty sure this is correct...

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    $\begingroup$ Welcome aboard. $\endgroup$ – Emre Apr 27 '11 at 5:52
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Try to express $s$ and $t$ as functions of $x$ and $y$ from the first two equations. Then plug these expressions into the third equation.

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