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? A. The system has two solutions, but only one is viable because the other results in a negative width. B. The system has only one solution, but it is not viable because it results in a negative width. C. The system has only one solution, and it is viable because it results in a positive width. D. The system has two solutions, and both are viable because they result in positive widths.
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? A. The system has two solutions, but only one is viable because the other results in a negative width. B. The system has only one solution, but it is not viable because it results in a negative width. C. The system has only one solution, and it is viable because it results in a positive width. D. The system has two solutions, and both are viable because they result in positive widths.
Solution
First, let's define the variables and write down the equations based on the problem:
Let x be the width of play area A. Then, the length of play area A is 4x + 1.
For play area B, the width is x + 2. And the length of play area B is 3(x + 2) + 2.
Since the areas of the two play areas are equal, we have the equation:
x * (4x + 1) = (x + 2) * [3(x + 2) + 2]
This simplifies to:
4x^2 + x = 3x^2 + 8x + 6
Rearranging terms gives us a quadratic equation:
x^2 - 7x - 6 = 0
Solving this quadratic equation will give us the possible values of x. The quadratic formula is x = [-b ± sqrt(b^2 - 4ac)] / (2a). In this case, a = 1, b = -7, and c = -6.
So, x = [7 ± sqrt((-7)^2 - 41(-6))] / (2*1) = [7 ± sqrt(49 + 24)] / 2 = [7 ± sqrt(73)] / 2
Since sqrt(73) is approximately 8.54, we have two possible solutions for x: (7 + 8.54) / 2 = 7.77 and (7 - 8.54) / 2 = -0.77.
However, since x represents a width, it cannot be negative. Therefore, the only viable solution is x = 7.77 feet.
So, the correct answer is A. The system has two solutions, but only one is viable because the other results in a negative width.
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