How do you solve #(2x+2)/(2x-1) = 1/3#?

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Answer 1 · 2016-03-08T11:23:37

#color(blue)(x=-3/4)#

Given:#" "color(brown)(1/3=(2x+2)/(2x-1))#

#color(green)("Multiply both sides by "color(blue)(3))#

#" "color(brown)(1xx(color(blue)(3))/3 =(color(blue)(3)(2x+2))/(2x-1)#

#color(green)("But "3/3 = 1" giving")#

#" "color(brown)(1=(6x+6)/(2x-1))#

#color(green)("Multiply both sides by "color(blue)((2x-1)))#

#" "color(brown)(1color(blue)(xx(2x-1)) = (6x+6)xx(color(blue)(2x-1))/(2x-1)#

#color(green)("But "(2x-1)/(2x-1)=1" giving")#

#" "color(brown)(2x-1=6x+6)#

#color(green)("Subtract "2x" from both sides giving:")#

#" " color(brown)(0-1=4x+6)->"Note that "6x-2x=4x#

#color(green)("Subtract 6 from both sides giving")#

#" "color(brown)( -7=4x)#

#color(green)("Divide bot sides by 4 giving")#

#" "color(brown)(-7/4=x)#

#" "color(blue)(x=-7/4)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check:

#(2x+2)/(2x-1) -> (2(-7/4)+2)/(2(-7/4)-1)= (-3/2)/(-9/2) =+ (3/2xx2/9) =3/9=1/3#

#color(red)("Confirmed as correct!")#