Seems Simple, forgetting some fundamentals - $\frac{1.6}{3.01} = \frac{x}{1000+x}$ I am having trouble remembering/finding simple source to review a few fundamentals. 
I know I need to factor and try getting x alone. Though not able to recreate answer. Seems it would have multiple roots. Would be great to see the proper steps to solve.
$$\frac{1.6}{3.01} = \frac{x}{1000+x}$$
 A: It does not have multiple roots, since the order of $x$ is just one. The multiplication happens between constants (numbers) so it does not affect the number of solutions.
$$\frac{1.6}{3.01}=\frac{x}{1000+x} \Leftrightarrow 1.6(1000+x)=3.01x \Leftrightarrow 1600 + 1.6x = 3.01x \Leftrightarrow 1.41x=1600 \Leftrightarrow \boxed{x = \frac{1600}{1.41}}$$
A: It's much easier than you think.
$\frac ab = \frac cd \iff ad = bc; b\ne 0; d\ne 0$.
So 
$\frac{1.6}{3.01} = \frac{x}{1000+x}\implies$
$(1000 + x)\frac{1.6}{3.01} = (1000 + x) \frac{x}{1000+x}\implies$
$(1000+x)\frac{1.6}{3.01}= x \implies$
$(1000+x)\frac{1.6}{3.01}*3.01 = x*3.01 \implies$
$1.6(1000 + x) = 3.01 x$
Just solve that.
(i.e.
$1600 + 1.6 x = 3.01 x$
$1600 = 3.01 x - 1.6 x = 1.41 x$
$x = \frac {1600}{1.41} $
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If you want to be pervese:
$\frac {1.6}{3.01} = \frac x{1000 + x}$
$\frac {160}{301} = \frac 1{\frac {1000}x + 1}$
$\frac {301}{160} = \frac {1000}x + 1$
$1\frac {141}{160}-1 = \frac {1000}x$
$\frac {141}{160} = \frac {1000}x$
$\frac {160}{141} = \frac x{1000}$
$\frac {160000}{141} = x$.
