SOLUTION: Two cyclists, 56 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of t

Question 831689: Two cyclists, 56 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

Found 2 solutions by hovuquocan1997, JulietG:
Answer by hovuquocan1997(83) (Show Source): You can put this solution on YOUR website!
Let the speed of the slow one be x
Let the speed of the fast one be y
Both travel and meet after 2 hours so we have the equation
2x + 2y = 56
Then you have the fast one is 3 times faster than the slow one, so you have
y = 3x =====>>>> 3x - y = 0
Then you solve the system of equation

Solved by pluggable solver: SOLVE linear system by SUBSTITUTION

Solve:
We'll use substitution. After moving 2*y to the right, we get:
, or . Substitute that
into another equation:
and simplify: So, we know that y=21. Since , x=7.

Answer: .

y=21 so the speed of the faster one is 21mi/h
TA-DAH
:D
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
A + 3A = 56/2 (two cyclists)
4A = 28
A = 7
.
The slower cyclist rides at 7mph. In 2 hours, he travels 14 miles.
The faster cyclist rides at 7*3mph. In 2 hours, he travels 42 miles.
Together, they travel 56 miles.
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