Question

How do you solve `1/x-1/(2x)=2x` ?

I know the solution to
`1/x-1/(2x)=2x`
is
`x=+-0.5`
At least it is the solution in my math book.

But I don't understand how you solve it.
When I tried i got `x=+-0,25` I think.

So I typed the problem into Cymath. And it tells me to do the following:

One. Simplify `1/x-1/(2x)` to `1/(2x)`

`1/(2x)=2x`

Two. Multiply both sides by `2x`

`1=2x*2x`

Three. Simplify `2x*2x` to `4x^2`

`1=4x^2`

Four. Divide both sides by `4`

`1/4=x^2`

Five. Take the square root of both sides.

`+-sqrt(1/4)=x`

So the answer is `x=+-0.5`

But the thing I don't understand is the first step. Why does `1/x-1/(2x)` equals to `1/(2x)`?

It just seems so illogical to me. Is there another way to solve the equation? Or can someone please explain why this is?

Answer 1

See the explanation please.

Explanation:

`1/x-1/(2x)=2x=>` Left side LCD = `2x` :

`2/(2x)-1/(2x)=2x =>` or:

`1/(2x)=2x`

Let's continue:

Multiply both sides by `2x`:

`(2x)/(2x)=2x*2x`

`1=4x^2 =>` divide both sides by`4`:

`1/4=x^2 =>` take square root of both sides:

`+-1/2=x =>` or:

`x=+-0.5`

2nd method:

`1/x-1/(2x)=2x`

To get rid of denominators multiply both sides of equation by the Lowest Common Denominator (LCD) in this case `2x`:

`(2x)/x-(2x)/(2x)=2x*2x =>` simplify:

`2-1=4x^2=>` simplify:

`1=4x^2 =>` continue same as above

Answer 2

`x=+-1/2`

Explanation:

Given: `1/x-1/(2x)=2x`

Considering the left hand side only for a moment `1/x-1/(2x)`
We need to make the bottom values (denominators) the same. Multiply by 1 and you do not change the value. However, 1 comes in many forms.

`color(green)([1/xcolor(red)(xx1)]-1/(2x)=2x color(white)("dddd")->color(white)("dddd")[1/xcolor(red)(xx2/2)]-1/(2x)=2x )`

`color(green)(color(white)("ddddddddddddddddddddd")->color(white)("ddddd")ubrace([2/(2x)]color(white)("dd")-1/(2x))=2x)`
`color(green)(color(white)("ddddddddddddddddddddddddddd-dddddd")darr)`

`color(green)(color(white)("ddddddddddddddddddddd")->color(white)("dddddddddd") 1/(2x)color(white)("dddd")=2x)`

Multiply both sides by `color(red)(x)`

`color(green)(color(white)("ddddddddddddddddddddd")->color(white)("ddddd") 1/(2cancel(x))color(red)(xx cancel(x))color(white)("dddd")=2xcolor(red)(xx x))`

`color(green)(color(white)("ddddddddddddddddddddd")->color(white)("ddddddddd")1/2=2x^2`

Divide both sides by 2

`color(green)(color(white)("ddddddddddddddddddddd")->color(white)("ddddddddd")1/4=x^2`

Square root both sides
`color(green)( color(white)("ddddddddddddddddddddd")->color(white)("ddddddd")+-1/2=x`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

Set `x=+1/2`
`1/x-1/(2x)=2x`

`1/(1/2)-1/(2xx1/2)=2xx1/2` ?

`2-1 = 1 color(red)(larr"True")`

Set `x=-1/2`
`1/x-1/(2x)=2x`

`1/(-1/2)-1/(2xx(-1/2))=2xx(-1/2)` ?

`-2+1 =- 1 color(red)(larr"True")`