# BASIC EQUATION

• Sep 28th 2008, 12:53 AM
caligyrl4lyfe
BASIC EQUATION
hey i was wondering if anyone could help me solve this equation for X.

a=1/1+x
• Sep 28th 2008, 01:12 AM
bkarpuz
Quote:

Originally Posted by caligyrl4lyfe
hey i was wondering if anyone could help me solve this equation for X.

a=1/1+x

It's so simple.
$a=\frac{1}{1+x}\text{ for }x\neq-1$, we have $x+1=\frac{1}{a}$, which gives $x=\frac{1}{a}-1=\frac{1-a}{a}$.

Note that $a=0$ if $|x|\to\infty$.
However, this can not hold for any finite value of $x$.
• Sep 28th 2008, 01:13 AM
Sean12345
Hi caligyrl4lyfe,

We have $a=\frac{1}{1+x}$ . Multiplying both sides by $1+x$ yields $a(1+x)=1$ . Now expanding the bracket gives $a+ax=1$ . Minusing the $a$ from both sides then dividing through by $a$ leads to $x=\frac{1-a}{a}$ . Which alternatively can be written as $x=\frac{1}{a}-1$ .

Hope this helps.
• Sep 28th 2008, 02:16 AM
Matt Westwood
Quote:

Originally Posted by bkarpuz
It's so simple.
$a=\frac{1}{1+x}\text{ for }x\neq-1$, we have $x+1=\frac{1}{a}$, which gives $x=\frac{1}{a}-1=\frac{a-1}{a}$.

Note that $a=0$ if $|x|\to\infty$.
However, this can not hold for any finite value of $x$.

Sorry dude, no, it's $\frac{1-a}{a}$.
• Sep 28th 2008, 02:18 AM
bkarpuz
Quote:

Originally Posted by Matt Westwood
Sorry dude, no, it's $\frac{1-a}{a}$.

(Rofl) yes! updating my post.