How do you solve #1/(2x-1) - 1/(2x+1) = 1/40 #?

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Answer 1 · 2016-02-08T01:30:19

#x=-9/2,x=9/2#

Get a common denominator on the left hand side of the equation.

#(2x+1)/((2x-1)(2x+1))-(2x-1)/((2x-1)(2x+1))=1/40#

Note that that the fractions can be combined and that the denominator is #(2x-1)(2x+1)=4x^2-1#.

#(2x+1-(2x-1))/(4x^2-1)=1/40#

#(2x+1-2x+1)/(4x^2-1)=1/40#

#2/(4x^2-1)=1/40#

Cross-multiply.

#80=4x^2-1#

#0=4x^2-81#

Factor this as a difference of squares.

#0=(2x+9)(2x-9)#

Now, set these both equal to #0# to find the two values of #x# that satisfy this equation.

#2x+9=0" "=>" "x=-9/2#

#2x-9=0" "=>" "x=9/2#