1. ## BASIC EQUATION

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

a=1/1+x

2. 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.
$\displaystyle a=\frac{1}{1+x}\text{ for }x\neq-1$, we have $\displaystyle x+1=\frac{1}{a}$, which gives $\displaystyle x=\frac{1}{a}-1=\frac{1-a}{a}$.

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

3. Hi caligyrl4lyfe,

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

Hope this helps.

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

Note that $\displaystyle a=0$ if $\displaystyle |x|\to\infty$.
However, this can not hold for any finite value of $\displaystyle x$.
Sorry dude, no, it's $\displaystyle \frac{1-a}{a}$.

5. Originally Posted by Matt Westwood
Sorry dude, no, it's $\displaystyle \frac{1-a}{a}$.
yes! updating my post.