Two airplanes leave the airport at the same time. One plane is flying 700mph at a bearing of N46°W and the other plane is flying 895mph at a bearing of N43° E. After 3 hours, how far are the planes from one another?

9514 1404 393

  3380 miles

Step-by-step explanation:

The distance separating the planes can be found using the law of cosines. One plane has gone 3×700 = 2100 miles; the other has gone 3×895 = 2685 miles. The angle between their flight paths is 46°+43° = 89°. Then the separation (s) is ...

  s² = 2100² +2685² -2(2100)(2685)cos(89°) = 11,619,225 -11,277,000×0.01745

  s² ≈ 11,422,414.21

  s ≈ √11,422,414.21 ≈ 3379.7

The planes are separated by about 3380 miles after 3 hours.

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.