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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?

Respuesta :

MrRoyal

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:

[tex]L_A \to[/tex] length of play area A

[tex]W_A \to[/tex] width of play area A

[tex]L_B \to[/tex] length of play area B

[tex]W_B \to[/tex] width of play area B

[tex]x \to[/tex] Area of A

[tex]y \to[/tex] Area of B

From the question, we have the following:

[tex]L_A = 1 + 4W_A[/tex]

[tex]W_B = 2 + W_A[/tex]

[tex]L_B = 2 + 3W_B[/tex]

[tex]x = y[/tex]

The area of A is:

[tex]x = L_A * W_A[/tex]

This gives:

[tex]x = (1 + 4W_A) * W_A[/tex]

Open bracket

[tex]x = W_A + 4W_A^2[/tex]

The area of B is:

[tex]y = L_B * W_B[/tex]

[tex]y = (2 + 3W_B) * ( 2 + W_A)[/tex]

Substitute: [tex]W_B = 2 + W_A[/tex]

[tex]y = (2 + 3(2 + W_A)) * ( 2 + W_A)[/tex]

Open brackets

[tex]y = (2 + 6 + 3W_A) * ( 2 + W_A)[/tex]

[tex]y = (8 + 3W_A) * ( 2 + W_A)[/tex]

Expand

[tex]y = 16 + 8W_A + 6W_A + 3W_A^2[/tex]

[tex]y = 16 + 14W_A + 3W_A^2[/tex]

We have that:

[tex]x = y[/tex]

This gives:

[tex]W_A + 4W_A^2 = 16 + 14W_A + 3W_A^2[/tex]

Collect like terms

[tex]4W_A^2 - 3W_A^2 + W_A -14W_A - 16 =0[/tex]

[tex]W_A^2 -13W_A - 16 =0[/tex]

Using quadratic calculator, we have:

[tex]W_A = -14.1[/tex] or [tex]W = 1.13[/tex] --- approximated

But the width can not be negative; So:

[tex]W = 1.19[/tex]

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