Question

select the correct answer. a carpenter is creating two new templates for his designs. one template will be in the shape of a right triangle, where the longer leg is 4 inches more than six times the shorter leg. the second template will be in the shape of a rectangle, where the width is 5 inches more than the triangles shorter leg, and the length is 3 inches. the carpenter needs the areas of the two templates to be the same. write a system of equations to represent this situation, where y is the area, and x is the length of the shorter leg of the triangle. which statement describes the number and viability of the systems solutions? a. the system has only one solution, and it is viable because it results in positive side lengths. b. the system has only one solution, but it is not viable because it results in negative side lengths. c. the system has two solutions, and both are viable because they result in positive side lengths. d. the system has two solutions, but only one is viable because the other results in negative side lengths.

Explanation:

Step1: Find the area of the right - triangle

The formula for the area of a right - triangle is \(y=\frac{1}{2}\times base\times height\). Let the shorter leg of the right - triangle be \(x\). Then the longer leg is \(6x + 4\). So the area of the right - triangle \(y=\frac{1}{2}x(6x + 4)=3x^{2}+2x\).

Step2: Find the area of the rectangle

The width of the rectangle is \(x + 5\) and the length is \(3\). The formula for the area of a rectangle is \(y=\text{length}\times\text{width}\). So the area of the rectangle \(y = 3(x + 5)=3x+15\).

Step3: Set up the system of equations

We have the system \(

$$\begin{cases}y=3x^{2}+2x\\y = 3x + 15\end{cases}$$

\). Set \(3x^{2}+2x=3x + 15\).

Step4: Solve the quadratic equation

Rearrange \(3x^{2}+2x-(3x + 15)=0\), which simplifies to \(3x^{2}-x - 15=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 3\), \(b=-1\), \(c=-15\)).
\(x=\frac{1\pm\sqrt{(-1)^{2}-4\times3\times(-15)}}{2\times3}=\frac{1\pm\sqrt{1 + 180}}{6}=\frac{1\pm\sqrt{181}}{6}\).
\(x_1=\frac{1+\sqrt{181}}{6}\approx\frac{1 + 13.45}{6}=\frac{14.45}{6}\approx2.41\) and \(x_2=\frac{1-\sqrt{181}}{6}\approx\frac{1-13.45}{6}\approx - 2.08\).

Since \(x\) represents the length of a side, \(x>0\). So \(x=\frac{1+\sqrt{181}}{6}\) is viable and \(x=\frac{1-\sqrt{181}}{6}\) is not (because side lengths cannot be negative).

Answer:

D. The system has two solutions, but only one is viable because the other results in negative side lengths.