1. Solving for a surd

(1+x)/x = sqrt3

I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

2. Originally Posted by 200001
(1+x)/x = sqrt3

I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

$1+x = x \sqrt{3}$

$1 = x\sqrt{3} - x$

$1 = x(\sqrt{3} - 1)$

$\frac{1}{\sqrt{3}-1} = x$

3. Hello 200001
Originally Posted by 200001
(1+x)/x = sqrt3

I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

You're right so far. Now solve this as you would any other equation: get the $x$'s on one side and everything else on the other.
$1+x = x\sqrt3$

$\Rightarrow 1 = x\sqrt3 -x$
Factorise:
$\Rightarrow 1 = x(\sqrt3-1)$
Divide:
$\Rightarrow x = \frac{1}{\sqrt3 -1}$
You can leave the answer like this, but it's neater if you 'rationalise the denominator':
$\Rightarrow x = \frac{1}{\sqrt3 -1}\color{red}\times \frac{\sqrt3+1}{\sqrt3+1}$
$= \frac{\sqrt3+1}{3-1}$

$=\frac{\sqrt3+1}{2}$

4. Both of you, thanks.....it was the final step that confused me

5. Originally Posted by 200001
(1+x)/x = sqrt3

I am struggling to get the hang of this. i have multiplied both sides by x and ended up with 1+ x = xsqrt3

Treat $\sqrt{3}$ like any other constant such as 4, 1 and -2.6
To do this make use of the difference of two squares $a^2-b^2=(a+b)(a-b)$
$\frac{1}{\sqrt{3} -1} \times \frac{\sqrt3 +1}{\sqrt3 +1} = \frac{\sqrt3 +1}{(\sqrt3 -1)(\sqrt3 + 1)} = \frac{\sqrt3 +1}{3-1} = \frac{\sqrt3 +1}{2}$