Question

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?

Answer 1

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:

$W = 1.19$