Question 149865
Hope you can follow this:
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((2x-a)/(x+a))-((x+3a)/(a-x))=(a^2/(a^2-x^2))+3 
factoring the right:
((2x-a)/(x+a))-((x+3a)/(a-x))=(a^2/((a-x)(a+x)))+3 
multiplying both sides by (a-x)(a+x):
((a-x)(2x-a))-((x+3a)(a+x)) = a^2+(3(a-x)(a+x)) 
expanding:
(2ax-2x^2-a^2+ax)-(ax+3a^2+x^2+3ax) = a^2+(3(a^2-x^2)) 
2ax-2x^2-a^2+ax-ax-3a^2-x^2-3ax = a^2 + 3a^2 - 3x^2 
-3x^2-4a^2-ax = a^2 + 3a^2 - 3x^2 
-ax = a^2 + 7a^2
0 = a^2 + -ax + 7a^2
0 = 8a^2 + -ax
ax = 8a^2
x = 8a