SOLUTION: Could you please hel me with this problem Solve for x 2over 3x + 2 over 3 = 8 over x+6 thanks

Question 30013: Could you please hel me with this problem
Solve for x
2over 3x + 2 over 3 = 8 over x+6
thanks
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
Solve for x
2over 3x + 2 over 3 = 8 over x+6
2/(3x)+2/3 = 8/(x+6)
Multiplying by 3x(x+6) (which is the lcm of 3,3x and (x+6) )
3(x+6)+2x(x+6)=8X(3x)
3x+18+2x^2+12x=24x
2x^2+(3x+12x-24x)+18 =0 (grouping like terms, changing sign while changing side)
2x^2-9x+18 =0 ----(*)
x = {-(-9) + or - sqrt[(-9)^2-4X2X18]}/2X2
( using formula x = [-b+or-sqrt(b^2-4ac)](2a) given ax^2+bx+c=0 )
= (1/4)[9+or-sqrt(81-144)]
=(1/4)[9+or-i(sqrt63)]
Answer: x= [9+i(sqrt(63)]/4 and x= [9-i(sqrt(63)]/4
Verification: x= [9+i(sqrt(63)]/4 in (1)
LHS = 2X[81-63+18i(sqrt(63)]/16 -9X[9+i(sqrt(63)]/4 +18
= [81-63+18i(sqrt(63)]/8-18X[9+i(sqrt(63)]/8+144/8
=(1/8)X[18+18i(sqrt(63)-162-18i(sqrt(63)+144]
=(1/8)X[(18-162+144)+0] = (1/8)X(0)=0 = RHS
Since complex roots always occur in conjugate pairs,
there is no need to test the validity of the other value

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