Question

Given that $x+y$ varies as $z+\frac{1}{z}$, and that $x-y$ varies as $z-\frac{1}{z}$, find the relation between $x$ and $z$, provided that $z=2$ when $x=3$ and $y=1$.

Answer 1

Let,

$x+y=k(z+\frac{1}{z})$

and $x-y=l(z-\frac{1}{z})$

Where $k.l$ are constants,

Now, for $z=2$ we have $x=3,y=1$ putting in first equation we have

$3+1=k(2+\frac{1}{2})$

$4=\frac{5}{2}k\Rightarrow k=\frac{8}{5}$

Putting in second equation we have,

$3-1=l(2-\frac{1}{2})$

$2=\frac{3}{2}l\Rightarrow l=\frac{4}{3}$

Adding both the equations we have,

$2x=k(z+\frac{1}{z})+l(z-\frac{1}{z})=\frac{8}{5}(z+\frac{1}{z})+\frac{4}{3}(z-\frac{1}{z})$

$2x=\frac{44}{15}z+\frac{4}{15z}$

$x=\frac{22}{15}z+\frac{2}{15z}$