It's way more complicated than that. The stopping won't be linear, because the kinetic energy of the two weights will not be linear. Whether you are overweight or underweight is irrelevant, as the distance is inversely proportional (assuming all other things are equal, including the overweight vehicle's ability to properly perform when overweight). If the weight is the same but the speed increased, doubling the speed quadruples the stopping distance. The same is true for mass, where if the speed is the same and only the mass gets doubled, the stopping distance is also quadrupled.
When you go from 6000 pounds to 9000 pounds, and increase of 50%, your stopping distance doubles, or is 200%. If you go from 6000 pounds to 12,000 pounds, an increase of 100%, your stopping distance quadruples, or is 400%.
A weight increase of 7.5% is minimal with regards to stopping distance, and even without doing the math, it's in the additional 10% range, plus or minus 5% for the speeds we travel (between 30 MPH and 75 MPH, anyway). Because of the increased kinetic energy at high speeds, the higher speeds like 65MPH and above will be more closer to the 15% further range, and the lower the speeds below 45MPH and it'll be closer to the 5% range. But 10% across the board is not a bad think to keep in mind.
So, if it takes you three car lengths to stop at 60 MPH, then it'll take you 3.3 car lengths with the extra weight.
Distance to stop, d = 0.5 * mass * velocity^2 / F
But that doesn't take into account braking pressure applied by the driver, temperature and pressure of the tires, the kinetic coefficient of the road, whether the road grade is slightly up or down hill, stuff like that.