Elia rode her bicycle from her house to the beach at a constant speed of 18 kilometers per hour, and then rode from the beach to the park at a constant speed of 15 kilometers per hour. The total duration of the rides was 1 hour and the distances she rode in each direction are equal.
Let b be the number of hours it took Elia to ride from her house to the beach, and p the number of hours it took her to ride from the beach to the park.
Which system of equations represents this situation?
Answers
Answer:
Answer:
Elia was riding \dfrac{5}{11}
11
5
of an hour from the house to the beach, \dfrac{6}{11}
11
6
of an hour from the beach to the house and rode
8\dfrac{2}{11}8
11
2
kilometers from the house to the beach and
8\dfrac{2}{11}8
11
2
kilometers from the beach to the house.
Step-by-step explanation:
1. Let b be the number of hours it took Elia to ride from her house to the beach, and p the number of hours it took her to ride from the beach to the park. The total duration of the rides was 1 hour, so
b + p = 1
2. Elia rode her bicycle from her house to the beach at a constant speed of 18 kilometers per hour, she was riding for b hour, then she rode 18b kilometers from her house to the beach.
Elia rode from the beach to the park at a constant speed of 15 kilometers per hour, she was riding for p hours, then she rode 15p kilometers from the beach to the house.
The distances she rode in each direction are equal, so
18b = 15p
3. Solve the system of two equations:
\begin{gathered}\left\{\begin{array}{l}b+p=1\\ \\18b=15p\end{array}\right.\end{gathered}
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⎪
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⎧
b+p=1
18b=15p
From the first equation
b=1-pb=1−p
Substitute it into the second equation
\begin{gathered}18(1-p)=15p\\ \\18-18p=15p\\ \\18=18p+15p\\ \\33p=18\\ \\p=\dfrac{18}{33}=\dfrac{6}{11}\\ \\b=1-\dfrac{6}{11}=\dfrac{5}{11}\end{gathered}
18(1−p)=15p
18−18p=15p
18=18p+15p
33p=18
p=
33
18
=
11
6
b=1−
11
6
=
11
5
Elia was riding \dfrac{5}{11}
11
5
of an hour from the house to the beach, \dfrac{6}{11}
11
6
of an hour from the beach to the house and rode
18\cdot \dfrac{5}{11}=\dfrac{90}{11}=8\dfrac{2}{11}18⋅
11
5
=
11
90
=8
11
2
kilometers to the beach and
15\cdot \dfrac{6}{11}=\dfrac{90}{11}=8\dfrac{2}{11}15⋅
11
6
=
11
90
=8
11
2
kilometers fro mthe beach to the house.