Solve for ##u## and ##v## in the given equations

[chwala]

Homework Statement

See attached.

Relevant Equations

Understanding of simultaneous equations

In my approach i have:

$u-v=\dfrac{1}{6(u+v)}$

$\dfrac{1}{u+v} + 12(u+v)=8$

$1+12(u+v)^2=8(u+v)$

Let

$u+v=m$

then we shall have,

$12m^2-8m+1=0$

$m_1=\dfrac{1}{2}$ and $m_2=\dfrac{1}{6}$

Using $m_2=\dfrac{1}{6}$ and considering
$(u+v)(u-v)=\dfrac{1}{6}$
then,
$\dfrac{1}{6} (u-v)=\dfrac{1}{6}$

then we shall have the simultaneous equation,

$u-v=1$
$u+v=\dfrac{1}{6}$ giving us
$u=\dfrac{7}{12} ⇒v=-\dfrac{5}{12}$

also using

$m_1=\dfrac{1}{2}$
then we shall have the simultaneous equation,
$u-v=\dfrac{1}{3}$
$u+v=\dfrac{1}{2}$ giving us
$u=\dfrac{5}{12} ⇒v=\dfrac{1}{12}$

There may be another approach hence my post. Cheers.

 

[haruspex]

Homework Statement: See attached.
Relevant Equations: Understanding of simultaneous equations

There may be another approach

I would have started $\frac 1{u+v}+\frac 2{u-v}=\frac{u-v+2(u+v)}{u^2-v^2}=(3u+v)6$.
No idea whether that is better.

 

[chwala]

I would have started $\frac 1{u+v}+\frac 2{u-v}=\frac{u-v+2(u+v)}{u^2-v^2}=(3u+v)6$.
No idea whether that is better.

@haruspex let me check that out...