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...

 

[chwala]

The other approach to this question though tedious

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

and noting that,

$3u-v=\dfrac{4}{3}$ from the first equation then,

$(u+\dfrac{4}{3}-3u)(u+\dfrac{4}{3}+3u)=\dfrac{1}{6}$

...

$\dfrac{72u-16-72u^2}{9}=\dfrac{1}{6}$

$-432u^2+432u-96=9$

$432u^2-432u+105=0$

$144u^2-144u+35=0$

$u =\dfrac{144±\sqrt {20736-20160}}{288}$

$= \dfrac{144±24}{288}$

$u_1=\dfrac{7}{12} ⇒v_1 = \dfrac{-5}{12}$

$u_2=\dfrac{5}{12} ⇒v_1 = \dfrac{1}{12}$

 

[Steve4Physics]

You are given $\frac 1{u+v} + \frac 2{u-v} = 8$ and $(u+v)(u-v) = \frac 16$. An apprach is...

Let $x = u+v$ and $y = u-v$. Also maybe use symbols for the constants: with $a=2, b=8,$ and $c = \frac 16$ the equations become:

$\frac 1x + \frac ay =b$ and $xy = c$.

From the first one $y + ax = bxy$. Since $xy = c$ this becomes $y + ax = bc$.

Substituting $y = bc~ – ~ax$ into $xy = c$ gives a quadratic in $x$ and the rest should follow.

Replace $a, b$ and $c$ when convenient.

Edit: typo's fixed.

 

[pasmith]

With x = 1/(u + v), y = 2/(u-v) we have <br /> x + y = 8 \quad xy = 12 so that x and y are the roots of <br /> z^2 - 8z + 3 = (z - 4)^2 - 4 = 0 giving z = 4 \pm 2 and hence \begin{split}<br /> (u,v) &amp;= \left( \frac12\left(\frac1x + \frac2y\right), \frac12\left(\frac1x - \frac2y\right)\right) \\<br /> &amp;= \left( \frac 7{12}, -\frac{5}{12} \right) \quad\mbox{or}\quad \left( \frac{5}{12}, -\frac{1}{12}\right)\end{split}

 

[Steve4Physics]

With x = 1/(u + v), y = 2/(u-v) we have <br /> x + y = 8 \quad xy = 12 so that x and y are the roots of <br /> z^2 - 8z + 3 = (z - 4)^2 - 4 = 0

I think there's a typo' and the last equation should be
$z^2 - 8z + 12 = (z - 4)^2 - 4 = 0$

giving z = 4 \pm 2 and hence \begin{split}<br /> (u,v) &amp;= \left( \frac12\left(\frac1x + \frac2y\right), \frac12\left(\frac1x - \frac2y\right)\right) \\<br /> &amp;= \left( \frac 7{12}, -\frac{5}{12} \right) \quad\mbox{or}\quad \left( \frac{5}{12}, -\frac{1}{12}\right)\end{split}

Maybe I’ve made a mistake but it seems $(u,v) = (\frac 5{12}, -\frac 1{12})$ is not a solution...

$u+v = \frac 5{12} + \frac {-1}{12} = \frac 4{12} = \frac 13$

$u-v = \frac 5{12} - \frac {-1}{12} = \frac 6{12} = \frac 12$

$\frac 1{u+v}+ \frac 2{u-v} = \frac 1{\frac 13} + \frac 2{\frac 12} = 3+ 4 =7$ (but it should be 8)

EDIT. I'm guessing it's just a sign-error so $(u,v) = (\frac 5{12}, -\frac 1{12})$ should be $(u,v) = (\frac 5{12}, \frac 1{12})$.

 

[SammyS]

With x = 1/(u + v), y = 2/(u-v) we have <br /> x + y = 8 \quad xy = 12 so that x and y are the roots of <br /> z^2 - 8z + 3 = (z - 4)^2 - 4 = 0 giving z = 4 \pm 2

Minor typo.

$\displaystyle z^2 - 8z + 3 = 0 \$ should be $\displaystyle z^2 - 8z + 12 = 0 \$

The rest follows from that without change.

 

[haruspex]

Minor typo.

$\displaystyle z^2 - 8z + 3 = 0 \$ should be $\displaystyle z^2 - 8z + 12 = 0 \$

The rest follows from that without change.

Except that there is also the sign error in post #6 that @Steve4Physics noted in post #7.

 

[Gavran]

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
$u=\dfrac{5}{12} ⇒v=\dfrac{1}{12}$

Good work.

 

[Ibix]

Defining $x=u+v$, $y=u-v$ the second equation is $xy=\frac 16$, or $6x=1/y$.

Substituting into the first gives $\frac 1{x}+12x=8$, which rearranges to $12x^2-8x+1=0$. Hence $x=1/2$ or $x=1/6$, with the corresponding $y=1/3$ or $y=1$.

Since $u$ and $v$ are obviously $\frac 12(x\pm y)$, then we have $(u,v)=\left(\frac{5}{12},\frac{1}{12}\right)$ or $\left(\frac{7}{12},\frac{-5}{12}\right)$.