# SOLUTION: Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:50 A.M. on his bicycle, traveling 10 mph faster than Kim. If they get to

Algebra ->  Algebra  -> Functions -> SOLUTION: Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:50 A.M. on his bicycle, traveling 10 mph faster than Kim. If they get to       Log On

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 Algebra: Functions, Domain, NOT graphing Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Functions Question 341165: Kim starts to walk 3 mi to school at 7:30 A.M. with a temperature of 0°F. Her brother Bryan starts at 7:50 A.M. on his bicycle, traveling 10 mph faster than Kim. If they get to school at the same time, then how fast is each one traveling? How do I solve this word problem because I am at a lost. I am totally pulling my hair out with this one. Please help me somebody!!!!Answer by galactus(183)   (Show Source): You can put this solution on YOUR website!I don't see what the temperature has to do with it other than it was a cold walk to school. Since d=rt, Kim travels 3=rt. Bryan starts 1/3 hour later and travels 10 mph faster, but he travels the same distance: 3=(r+10)(t-1/3) There are two equations with two unknowns. Can you finish now?.