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 $y$.
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 $y$.
Hint: What is $\dfrac{x}{y}$ equal to?
$x=s^{2}-t^{2}=(s+t)(s-t)$ 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...
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.