At Glencliff High School, the photography
2 answers:
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Answer:
А.The system has two solutions, but only one is viable because the other results in a negative width.
Step-by-step explanation:
Given
Let:
length of play area A
width of play area A
length of play area B
width of play area B
Area of A
Area of B
From the question, we have the following:
The area of A is:
This gives:
Open bracket
The area of B is:
Substitute:
Open brackets
Expand
We have that:
This gives:
Collect like terms
Using quadratic calculator, we have:
or
--- approximated
But the width can not be negative; So: