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