How do you solve #(6/(x^2-1)) - (1/(x+1)) = 1/2#?

Question

Eva Abramson · Accepted Answer

#x~~ -4.464" or "2.464# to 3 decimal paces

#x^2-1# can be written as #x^2-1^2# Compare this to #a^2+b^2=(a+b)(a-b)# We can use this to our advantage Write the given equation as: #6/(x^2-1^2) - 1/(x+1)=1/2# By substitution this is #6/((x+1)(x-1)) - 1/(x+1)=1/2# #(6-(x-1))/((x+1)(x-1))=1/2# Multiply both sides by 2 #(12-2(x-1))/((x+1)(x-1))=1# #12-2(x-1)=(x+1)(x-1)# #12-2x+2=x^2-1# #x^2+2x-11=0# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Now solve as a quadratic - Completing the square method (skipping many steps) #(x+1)^2 -12=0# #x+1= +-sqrt(12)# #x= -1+-2sqrt(3)# #x~~ -4.464" or "2.464# to 3 decimal paces

Nathan Axel · Answer

undefined To solve the equation (6/(x^2-1)) - (1/(x+1)) = 1/2, we can follow these steps:

1. Find a common denominator for the fractions on the left side of the equation. The common denominator is (x^2-1)(x+1).

2. Multiply each term by the common denominator to eliminate the fractions.

3. Simplify the equation by distributing and combining like terms.

4. Rearrange the equation to isolate the variable, x.

5. Solve the resulting quadratic equation by factoring or using the quadratic formula.

6. Check the solutions obtained by substituting them back into the original equation to ensure they satisfy the equation.

Please let me know if you need further assistance.