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

Question

8561 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-31T06:26:00

$x=pi/(pi-1)$

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.

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