I have done many problems of this type and tutor students who leave parenthesis off when doing test. Unfortunately if you do it on a test it results in wrong answers and a low grade. To me the problem is
(x-2)/(x+4) + (x+1)/(x+6) = (11x+32)/(x2+10x+24)
Step 1) Factor all denominators and find the restrictions on x. (x+4) and (x+6) are already factored. x2+10x+24 = (x+4)(x+6)
(x-2)/(x+4) + (x+1)/(x+6) = (11x+32)/((x+4)(x+6))
We know we cannot divide by zero. We see from above that x cannot equal -4 and -6.
Step 2) Find the LCD. We see that the LCD = (x+4)(x+6).
Step 3) Multiply each term by the LCD. This will clear all denominators.
(x-2)(x+4)(x+6)/(x+4) + (x+1)(x+4)(x+6)/(x+6) = (11x+32)(x+4)(x+6)/((x+4)(x+6))
Step 4) Cancel.
(x-2)(x+6) + (x+1)(x+4) = 11x + 32
Step 5) Solve for x by expanding and simplifying.
x2 + 4x -12 +x2 +5x + 4 = 11x + 32
2x2 + 13x - 8 = 11x + 32 Moving all terms to the left
2x2 +2x - 40 = 0 Dividing each term by 2
x2 + x - 20 = 0
Since this is a quadratic equation we can either factor or use the quadratic formula.
(x + 5)(x - 4) = 0
x + 5 = 0 → x = -5 or x - 4 = 0 → x = 4
Step 6) Check your solutions to make sure that none are extraneous.
Since x does not equal -4 or -6 we see that both solutions are valid.