How do you solve 2/x+2 + 1/x-2 = 13/21 and 2x-5/x-3 -2 = 3/x+3

2 +1 = 13
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x+2 x-2 21

using 21(x-2)(x+2) as the lcd we have

2(21)(x-2)+ 1(21)(x+2)= 13(x-2)(x+2)
2(21x-42+ 1(21x+42)= 13x^2-52
42x-84+21x+42=13x^2-52
63x-42=13x^2-52
13x^2-63x-10=0

(13x+2)(x-5)=0
x=-2/13 or 5

2x-5 -2 = 3
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x-3 1 x+3

(2x-5)(x+3)-2(x^2-9)=3(x-3)

2x^2+x-15-2x^2+18=3x-9
x+3=3x-9
12=2x
6=x

(2/x+2) + (1/x-2) = 13/21. I am assuming these are set up as fractions. If they are not, do so. Ex. 2 OVER x+2. Then you want to find the least common denominator which appears to be 21(x+2)(x-2). Then you will do whats needed to get each fractions to the same denominator. Example: 2 over x+2, to get x+2 to the correct denominator, you multiply it by 21(x-2). What ever you do to the denominator, you do to the numerator. Then, you treat it like a normal addition problem once you get everything over the correct denominator. Same with the other one, hope i helped:)

1).
2/x+2 + 1/x-2 = 13/21
21*2(x - 2) + 21*1(x + 2) = 13(x^2 - 4)
13x^2 - 63x - 52 + 84 - 42 = 0
13x^2 - 63x - 10 = 0
(13x + 2)(x - 5) = 0
Solutions:
x = -2/13
x = 5

2).
2x-5/x-3 -2 = 3/x+3
(2x - 5)(x + 3) - 2(x^2 - 9) = 3(x - 3)
2x - 12 = 0
2(x - 6) = 0
Solution:
x = 6