(3x-1)+{1/(3x-1)}=5

Mutiply by the denominator (3x-1) to get rid of the fractional form.

(3x-1)^2+1=5(3x-1)

9x^2-6x +2=15x-5.

9x^2-6x-15x+7=0

9x^2-21x+7 is a quadratic equation of the form ax^2+bx+c=0 whose roots are given by

x=[b-sqrt(b^2-4ac)}/(2a).

a=9,b=-21 and c=7. Substing in the quadratic equation we get the roots:

x= [-(-21) +or- sqrt(21^2-4*9*7)}/(2*9}

**x=[7+sqrt21]6 or x=[7-sqrt21]/6**

**x=1.9304**approximately, or **x=0.4029 **approx.

You can put back int he original equations

(3x-1)+[1/(3x-1)]=5

(3x-1)(3x-1) + 1 = 5(3x-1)

9x^2 -6x + 1 +1=15x-5

9x^2 -21x +2 = - 5

9x^2 -21x +7 = 0

( )( )=0

check the equation--it doesn't factor--see if you wrote it down right please-- I think you meant = negative 5--then it would work fine

then send it back to me . . .I love doing these things

thanks