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disa [49]
2 years ago
9

A school is planning to construct two rectangular play areas in the playground. The length of play area A must be 1 foot longer

than four times its width. The width of play area B must be 2 feet longer than the width of play area A, and the length must be 2 feet longer than three times its own width. In addition, the areas of the two play areas must be equal. Write a system of equations to represent this situation, where y is the area of the play areas and x is the width of play area A. Which statement describes the number and viability of the system's solutions?
Mathematics
1 answer:
Anna11 [10]2 years ago
7 0

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:

L_A \to length of play area A

W_A \to width of play area A

L_B \to length of play area B

W_B \to width of play area B

x \to Area of A

y \to Area of B

From the question, we have the following:

L_A = 1 + 4W_A

W_B = 2 + W_A

L_B = 2 + 3W_B

x = y

The area of A is:

x = L_A * W_A

This gives:

x = (1 + 4W_A) * W_A

Open bracket

x = W_A + 4W_A^2

The area of B is:

y = L_B * W_B

y = (2 + 3W_B) * ( 2 + W_A)

Substitute: W_B = 2 + W_A

y = (2 + 3(2 + W_A)) * ( 2 + W_A)

Open brackets

y = (2 + 6 + 3W_A) * ( 2 + W_A)

y = (8 + 3W_A) * ( 2 + W_A)

Expand

y = 16 + 8W_A + 6W_A + 3W_A^2

y = 16 + 14W_A + 3W_A^2

We have that:

x = y

This gives:

W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2

Collect like terms

4W_A^2 - 3W_A^2 + W_A  -14W_A  - 16 =0

W_A^2  -13W_A  - 16 =0

Using quadratic calculator, we have:

W_A = -14.1 or W = 1.13 --- approximated

But the width can not be negative; So:

W = 1.19

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An online furniture store sells chairs for $150 each and tables for $550 each. Every day, the store can ship a maximum of 40 pie
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Answer:

minimum of 13 chairs must be sold to reach a target of $6500

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Step-by-step explanation:

Given that:

Price of chair = $150

Price of table = $400

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According to given condition:

t + c = 30 ----------- eq1

t(150) + c(400) = 6500 ------ eq2

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Putting in eq1

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