# Two difficult solve for x qns.

• Jan 6th 2010, 05:01 PM
-Equilibrium-
Two difficult solve for x qns.
Working on this set of questions, and Ive come to two difficult qns, which Im not sure which route I should take at tackling them, its come to a point where Ive written and erased it so many times.. my page has ripped (Angry) *pressure*
Any help would be greatly appreciated.
Solve for x:
1.
x/m + n = x/n + m

2.
(1/ x + a) + (1 / x + 2a) = 2 /x + 3a

Thank you. (Hi)
• Jan 6th 2010, 06:11 PM
vonflex1
if i understand the questions correctly:

$\displaystyle \frac{x}{m}+n = \frac{x}{n}+m$

$\displaystyle x+nm = x+nm$

$\displaystyle 0 = 0$

which means this equation is true for every x...

and the second one is similar:

$\displaystyle (\frac{1}{x}+a)+(\frac{1}{x}+2a) = \frac{2}{x}+3a$

$\displaystyle \frac{2}{x}+3a = \frac{2}{x}+3a$

which also means this equation is true for every x, except x=0,
when $\displaystyle \frac{1}{x}$ is not defined.
• Jan 6th 2010, 08:27 PM
dkaksl
How did you simplify the first one, vonflex1?

I got

$\displaystyle \frac{x}{m}+n=\frac{x}{n}+m$

$\displaystyle \frac{x}{m}-\frac{x}{n}=m-n$

$\displaystyle \frac{xn-xm}{mn}=m-n$

$\displaystyle \frac{x(n-m)}{mn}=m-n$

$\displaystyle x=\frac{mn(m-n)}{n-m}$

but I might just be confusing myself somewhere
• Jan 6th 2010, 08:36 PM
Stroodle
Quote:

Originally Posted by dkaksl
How did you simplify the first one, vonflex1?

I got

$\displaystyle \frac{x}{m}+n=\frac{x}{n}+m$

$\displaystyle \frac{x}{m}-\frac{x}{n}=m-n$

$\displaystyle \frac{xn-xm}{mn}=m-n$

$\displaystyle \frac{x(n-m)}{mn}=m-n$

$\displaystyle x=\frac{mn(m-n)}{n-m}$

but I might just be confusing myself somewhere

You're correct, but you can simplify it further:

$\displaystyle x=\frac{mn(m-n)}{n-m}$

$\displaystyle x=\frac{mn(m-n)}{-(m-n)}$

$\displaystyle x=-mn$
• Jan 6th 2010, 09:24 PM
vonflex1
• Jan 7th 2010, 02:29 AM
-Equilibrium-
any ideas about the second question?
cos the answer for the first one is right :) thank you
• Jan 7th 2010, 03:17 AM
vonflex1
did i write that second question right in latex?

is this the exact question or is there a

$\displaystyle x+2a, x+a, x+3a$ in the denominator instead of just the $\displaystyle x$?
• Jan 7th 2010, 03:24 AM
nikhil
[1/(x+a)]+[1/(x+2a)]=2/(x+3a)
this is equal to
(x+a+x+2a)/[(x+a)(x+2a)]= 2/(x+3a)
=(2x+3a)/[(x+a)(x+2a)]= 2/(x+3a)
now cross multiply and we get
2x^2+9ax+9a^2=2x^2+6ax+4a^2
or
3ax=-5a^2
finally
x=-5a/3

2) or on the second thought if your question is
[(1/x)+a]+[(1/x)+2a]=[(2/x)+3a]
then it is an identity. x can have any value except 0