Seems Simple, forgetting some fundamentals - $\frac{1.6}{3.01} = \frac{x}{1000+x}$

Question

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.

$$\frac{1.6}{3.01} = \frac{x}{1000+x}$$

Answer 1

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.

$$\frac{1.6}{3.01}=\frac{x}{1000+x} \Leftrightarrow 1.6(1000+x)=3.01x \Leftrightarrow 1600 + 1.6x = 3.01x \Leftrightarrow 1.41x=1600 \Leftrightarrow \boxed{x = \frac{1600}{1.41}}$$

Answer 2

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} $

=====

If you want to be pervese:

$\frac {1.6}{3.01} = \frac x{1000 + x}$

$\frac {160}{301} = \frac 1{\frac {1000}x + 1}$

$\frac {301}{160} = \frac {1000}x + 1$

$1\frac {141}{160}-1 = \frac {1000}x$

$\frac {141}{160} = \frac {1000}x$

$\frac {160}{141} = \frac x{1000}$

$\frac {160000}{141} = x$.