Ew. Step one is to clear all those yucky fractions. We need to find the smallest number that 3, 4, and 12 can all go into evenly.
3, 6, 9, 12
4, 8, 12
6, 12
12
Whatever we do to the denominators we need to do to the numerators. 3*4 is 12, so 2*4 is 8. 6*2 is 12, so 1*2 is 2. 4*3 is 12, so 1*3 is 3. The final one stays the same because the denominator is already 12.
Now just solve.
[tex] \frac{8v}{12} + \frac{2v}{12} - \frac{3v}{12} = \frac{7}{12} [/tex]
8v+2v-3v=7.
7v=7.
v=1.
