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1 answer:
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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:
Answer:
Step-by-step explanation:
let the original fraction be
( with denominator 1 more than numerator )
Subtracting 3 from numerator/ denominator gives
=
Given that the fraction is now equivalent to , then
=
( cross- multiply )
2(n - 3) = n - 2, distribute left side
2n - 6 = n - 2 ( subtract n from both sides )
n - 6 = - 2 ( add 6 to both sides )
n = 4
and n + 1 = 4 + 1 = 5
Thus original fraction is