Question

How do you solve `2/( 5x) = 35x`?

Answer 1

`x=+-sqrt14/35`

Explanation:

First we multiply both sides by `5x`:
`2/cancel(5x)*cancel(5x)=35x*5x`

`2=175x^2`

Divide both sides by `175`:
`(cancel(175)x^2)/cancel175=2/175`

Take the square root of both sides:
`sqrt(x^2)=+-sqrt(2/175)`

`x=+-sqrt(2/175)`

`x=+-sqrt(2)/sqrt(5*5*7)`

`x=+-sqrt2/(5sqrt7)`

`x=+-(sqrt2*sqrt7)/(5sqrt7*sqrt7)`

`x=+-sqrt14/35`

Answer 2

`x=sqrt14/35`

Explanation:

To 'get rid of' the `5x` from `2/(5x)` multiply both sides by `5x` giving

`color(green)(2/(cancel(5x))color(red)(xxcancel(5x))=35xcolor(red)(xx5x))`

`color(green)(2=175x^2)`

To 'get rid of' the `175" from "175x^2` multiply each side by `1/175`

`color(green)(2 color(red)(xx1/175) color(white)("d")=color(white)("d")x^2 xxcancel(175)color(red)(xx1/cancel(175)))`

`x=+-sqrt(2/175)`

`x=+-sqrt2/sqrt(5^2xx7)color(white)("dd")=color(white)("dd")+-sqrt2/(5sqrt7)`

Not good practice to have a root in the denominator so lets 'get rid' of it. Multiply by 1 and you do not change the value. However, 1 comes in many forms

`color(green)(x=+-sqrt2/(5sqrt7)color(red)(xx1)color(white)("d")=color(white)("d")+-sqrt2/(5sqrt7)color(red)(sqrt7/sqrt7))`

`x=+-sqrt14/35`