Question

Given `(2x)/(4pi) + (1-x)/2 = 0`, how do you solve for x?

Answer 1

`x=pi/(pi-1)`

Explanation:

The given equation:

`(2x)/(4pi)+(1-x)/2=0`

Multiply both sides of the equation by `4pi`

`(4pi)*[(2x)/(4pi)+(1-x)/2]=(4pi)*0`

`[(2x)+(2pi)(1-x)]=0`

`2x+2pi-2pi*x=0`

`(2-2pi)x=-2pi`

Divide both sides of the equation by `(2-2pi)`

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

`(cancel((2-2pi))x)/cancel((2-2pi))=(-2pi)/(2-2pi)`

`x=(-2pi)/(2-2pi)" "->" "x=(2(-pi))/(2(1-pi))`

Divide every term by 2 in both numerator and denominator

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

`x=pi/(pi-1)`

God bless....I hope the explanation is useful.