# Thread: Find in terms of C

1. ## Find in terms of C

Cheers

Find x in terms of c, given that ( 2 / 3x - 2c ) + (3 / 2x - 3c) = 7 / 2c

2. just a question

$\displaystyle (\frac{2}{3x}-2c)+(\frac{3}{2x}-3c)=\frac{7}{2c}$

or

$\displaystyle (\frac{2}{3x-2c})+(\frac{3}{2x-3c})=\frac{7}{2c}$

3. this is the exact question, sorry about the inital setout.

4. Originally Posted by Joel

this is the exact question, sorry about the inital setout.
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.