BASIC EQUATION

**caligyrl4lyfe** · September 28th 2008, 12:53 AM

hey i was wondering if anyone could help me solve this equation for X.

a=1/1+x

**bkarpuz** · September 28th 2008, 01:12 AM

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 $x$ .

**Sean12345** · September 28th 2008, 01:13 AM

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.

**Matt Westwood** · September 28th 2008, 02:16 AM

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 $\frac{1-a}{a}$ .

**bkarpuz** · September 28th 2008, 02:18 AM

Originally Posted by Matt Westwood

Sorry dude, no, it's $\frac{1-a}{a}$ .

yes! updating my post.

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