SOLUTION: Jack left school and walked home down Snakey Lane. His father left home at the same time and walked up Snakey Lane. If his father is walking twice as fast as Jack, how far will h

Question 350506: Jack left school and walked home down Snakey Lane. His father left home at the same time and walked up Snakey Lane. If his father is walking twice as fast as Jack, how far will he walk before they meet. The distance between Jacks house and the school, on Skakey Lane, is 2 miles.
Answer by jjordan95(63) (Show Source): You can put this solution on YOUR website!
You can create an algebraic expression out of this to solve for the distance.
Jack =
Jack's father = (He walks at 2 times the speed of Jack)
miles <--- Original
miles <--- Simplify
of a mile <--- Solve for Jack's speed
Therefore, Jack's distance is of a mile
Based on the original equation,
<--- By moving Jacks distance to the right side, you can now solve for his father's distance
<---Plug in Jack's distance
<-- Common denominators
<--- Answer
Therefore, the distance Jack's father walks is of a mile.
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