How do you solve #1/(x+3) + 1/(x-3) = 1/(x^2-9)#? Algebra Rational Equations and Functions Clearing Denominators in Rational Equations 1 Answer Don't Memorise May 27, 2015 #color(blue)(a^2 - b^2 = (a+b)(a-b)# #(x^2-9) = (x+3)(x-3)# the expression given is: #1/(x+3) + 1/(x-3) = 1/(x^2-9)# #1/(x+3) + 1/(x-3) = 1/((x+3)(x-3)# #(1.(x-3))/((x+3)(x-3)) + (1.(x+3))/((x-3)(x+3)) = 1/((x+3)(x-3)# #((x-3)+ (x+3))/((x+3)(x-3)) = 1/((x+3)(x-3))# #(2x)/cancel(x^2-9) = 1/cancel(x^2-9)# #2x = 1# # x =color(blue)( 1/2# is the solution for the expression. Answer link Related questions What is Clearing Denominators in Rational Equations? How do you solve rational expressions by multiplying by the least common multiple? How do you solve #5x-\frac{1}{x}=4#? How do you solve #-3 + \frac{1}{x+1}=\frac{2}{x}# by finding the least common multiple? What is the least common multiple for #\frac{x}{x-2}+\frac{x}{x+3}=\frac{1}{x^2+x-6}# and how do... How do you solve #\frac{x}{x^2-36}+\frac{1}{x-6}=\frac{1}{x+6}#? How do you solve by clearing the denominator of #3/x+2/x^2=4#? How do you solve #2/(x^2+2x+1)-3/(x+1)=4#? How do you solve equations with rational expressions #1/x+2/x=10#? How do you solve for y in #(y+5)/ 2 - y/3 =1#? See all questions in Clearing Denominators in Rational Equations Impact of this question 1440 views around the world You can reuse this answer Creative Commons License