You can put this solution on YOUR website! 5/x-(2x)/(x+2)=(x^2+10x+8)/(x^2+2x) ----(1)
5/x-(2x)/(x+2)=(x^2+10x+8)/[x(x+2)]
Multiplying byx(x+2) through out
5(x+2)-(2x)X(x)= (x^2+10x+8)
5x+10-2x^2 = x^2+10x+8
0 = (x^2+2x^2)+(10x-5x)+(8-10) (grouping like terms)
0 = 3x^2+5x-2
That is 3x^2+5x-2 =0
3x^2+(6x-x)-2 =0
( sum is 5 and the product is -6 and so the quantities are +6 and (-1)
(3x^2+6x)-x-2 =0 (by additive associativity)
3x(x+2)-1(x+2) =0
3xp-p= 0 where p = (x+2)
p(3x-1) = 0
(x+2)(3x-1) = 0
(x+2) = 0 gives x= -2
(3x-1) = 0 gives x =1/3
But x = -2 implies (x+2) = 0 and since division by zero is not defined therefore x = -2 DOES NOT HOLD
Verification: Now for x = 1/3 in (1)
LHS = 5/x-(2x)/(x+2)
=15 -[(2/3)/(7/3)]
=15-2/7 = 103/7
RHS=(x^2+10x+8)/(x^2+2x)= [(1/9)+(10/3)+8]/(1/9+2/3)
=[(1+30+72)/9]/(7/9)
=103/7 =LHS
Therefore, Answer: x = 1/3
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