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

Question

Kennedy Adams · Accepted Answer

#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#

Eva Abramson · Answer

undefined To solve the equation 1/(2x-1) - 1/(2x+1) = 1/40, you can follow these steps:

1. Find a common denominator for the fractions on the left side of the equation. In this case, the common denominator is (2x-1)(2x+1).

2. Multiply each term by the common denominator to eliminate the fractions. This gives us (2x+1) - (2x-1) = (2x-1)(2x+1) * (1/40).

3. Simplify the equation by distributing and combining like terms. This results in 2x + 1 - 2x + 1 = (2x^2 - 1) * (1/40).

4. Further simplify the equation by combining like terms. This gives us 2 = (2x^2 - 1) * (1/40).

5. Multiply both sides of the equation by 40 to eliminate the fraction. This yields 80 = 2x^2 - 1.

6. Move all terms to one side of the equation to set it equal to zero. This gives us 2x^2 - 81 = 0.

7. Factor the quadratic equation or use the quadratic formula to solve for x. In this case, the equation can be factored as (2x - 9)(x + 9) = 0.

8. Set each factor equal to zero and solve for x. This gives us two possible solutions: x = 9/2 or x = -9.

Therefore, the solutions to the equation 1/(2x-1) - 1/(2x+1) = 1/40 are x = 9/2 and x = -9.