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

• chwala
chwala
Gold Member
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.

chwala said:
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
haruspex said:
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...

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