Question 1123436
{{{ (2x)/(4x-1)+(x)/(4x+1)=(1)/(16x^2-1)+(1)/(2)}}}


{{{ (2x(4x+1)+x(4x-1))/((4x+1)(4x-1))=(16 x^2 + 1)/(2 (16 x^2 - 1))}}}


{{{ (8x^2+2x+4x^2-x)/((16 x^2 - 1))=(16 x^2 + 1)/(2 (16 x^2 - 1))}}}


{{{ (12x^2+x)/((16 x^2 - 1))=(16 x^2 + 1)/(2 (16 x^2 - 1))}}}


{{{ 2 (16 x^2 - 1)(12x^2+x)=(16 x^2 + 1)(16 x^2 - 1)}}}


{{{ 2* cross((16 x^2 - 1))(12x^2+x)=(16 x^2 + 1)cross((16 x^2 - 1))}}}


{{{ 2 (12x^2+x)=16 x^2 + 1}}}


{{{ 24x^2+2x=16 x^2 + 1}}}


{{{ 24x^2+2x-16 x^2 - 1=0}}}


{{{ 8x^2+2x - 1=0}}}


{{{ 8x^2 -2x+4x- 1=0}}}


{{{ (8x^2 -2x)+(4x- 1)=0}}}


{{{ 2x(4x -1)+(4x- 1)=0}}}


{{{(2x + 1) (4x - 1) = 0}}}


solutions:

if {{{(2x + 1) = 0}}}->{{{2x=-1}}}->{{{highlight(x=-1/2)}}}=>real solution

if {{{ (4x - 1) = 0}}}->{{{4x=1}}}->{{{x=1/4}}}-> since {{{ (4x - 1)}}} is denominator, it's not solution; so {{{x<>1/4}}}