Having trouble with making the common denominator work. please help.
Cheers
Find x in terms of c, given that ( 2 / 3x - 2c ) + (3 / 2x - 3c) = 7 / 2c
Then don't worry about "common denominator" or adding fractions: multiply the entire equation by (3x-2c)(2x- 3c)(2c) (which is, actually, the "common denominator.)
Multiplying the first term by that, $\displaystyle \frac{2}{3x-2c}(3x-2c)(2x-3c)(2c)$, the "3x- 2c" terms cancel leaving $\displaystyle 2(2x-3c)(2c)= 4c(2x-3c)= 8cx- 12c^2$. Multiplying the second term by that, $\displaystyle \frac{3}{2x-3c}(3x-2c)(2x-3c)(2c)$, the "2x-3c" terms cancel leaving $\displaystyle 3(2c)(2x-3c)= 6c(2x-3c)= 12cx- 18c^2$. Finally, multiplying the right hand side of the equation by that, $\displaystyle \frac{7}{2c}(3x-2c)(2x-3c)(2c)$, the "2c" terms cancel leaving $\displaystyle 7(3x-2c)(2x-3c)= 7(6x^2-10cx+ 6c^2)= 42x^2- 70cx+ 42c^2$.
Putting those together, multiplying the entire equation by (3x-2c)(2x-3c)(2c) changes the equation to $\displaystyle 8cx- 12c^2+ 12cx- 18c^2= 42x^2- 70cx+ 42c^2$ or $\displaystyle 20cx- 30c^2= 42x^2- 70cx+ 42c^2$, $\displaystyle 42x^2- 90cx+ 72c^2= 0$. Dividing through by 2, $\displaystyle 21x^2- 45cx+ 36c^2= 0$.
That appears, to me, to not have any real roots.