Factor the first denominator by factoring out 3 The LCD is 6(x-2) so we multiply every term through by We cancel whatever will cancel: What we have left is: Get 0 on the right: Multiply through by -1 so x2 term will be positive: Factor: 5x-2 = 0; x-2 = 0 5x = 2 x = 2 x = Two answers are and . However, the original problem has denominators that contain variables. Therefore we must check for extraneous answers that are not solutions because they cause the denominator to equal 0. No denominator can ever equal to 0. The denominators in the original equation are 3x-6 and x-2 Substituting for x in denominator 3x-6 is which is not 0 Substituting for x in denominator x-2 is which is not 0 either, so is a solution.. Substituting for x in denominator 3x-6 is so we must discard the answer x = 2, for it is extraneous and is not a solution. Therefore there is only one solution, x = ---------------------------------------- The second one: Add to both sides: Since both terms on the right have the same denominator we can just add the numerators and place it over their common denominator: Simplify the right side by dividing top and bottom by 4: We could get an LCD as we did in the other one but since there is only a fraction on each side, we can just cross-multiply: Get 0 on the right: 3x+2 = 0; x-2 = 0 3x = -2; x = 2 x = Two answers are and . However, the original problem also has denominators that contain variables. Therefore we must check for extraneous answers that are not solutions because they cause the denominator to equal 0. No denominator can ever equal to 0. The denominators in the original equation are x-1 and 2x. Substituting for x in denominator x-1 is which is not 0 Substituting for x in denominator 2x is which is not 0 either, so is a solution. Substituting for x in denominator x-1 is which is not 0 Substituting for x in denominator 2x is which is not 0 either, so is also a solution. So there are two solutions and 2. But as you see we must make sure that any answer we get really is a solution by substituting into each denominator to show that no answer causes any denominator in the original equation to be 0. Edwin