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.

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

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

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