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

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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.