Problem 18 - Precalculus Student Solutions Manual 5th

Question

Solve each problem. Concept Check Which one or more of the following cannot be a correct equation to solve a geometry problem, if \(x\) represents the length of a rectangle? (Hint: Solve each equation and consider the solution.) A. \(2 x+2(x-1)=14\) B. \(-2 x+7(5-x)=52\) C. \(5(x+2)+5 x=10\) D. \(2 x+2(x-3)=22\)

Step-by-Step Solution

Answer

Options B and C cannot be correct equations, as their solutions are not valid lengths.

Step 1 - Solve equation A

Solve the equation for option A:\(\begin{aligned}2x + 2(x-1) &= 14\ 2x + 2x - 2 &= 14\ 4x - 2 &= 14\ 4x &= 16\ x &= 4ewline\text{The solution is } x = 4\text{, which is a valid length.} \end{aligned}\)

Step 2 - Solve equation B

Solve the equation for option B:\(\begin{aligned}-2x + 7(5-x) &= 52\ -2x + 35 - 7x &= 52\ -9x + 35 &= 52\ -9x &= 17\ x &= -\frac{17}{9}ewline\text{The solution is } x = -\frac{17}{9}\text{, which is not valid because lengths cannot be negative.} \end{aligned\)}

Step 3 - Solve equation C

Solve the equation for option C:\(\begin{aligned}5(x+2) + 5x &= 10\ 5x + 10 + 5x &= 10\ 10x + 10 &= 10\ 10x &= 0\ x &= 0ewline\text{The solution is } x = 0\text{, which is not a valid length as length cannot be zero.} \end{aligned}\)

Step 4 - Solve equation D

Solve the equation for option D:\(\begin{aligned}2x + 2(x-3) &= 22\ 2x + 2x - 6 &= 22\ 4x - 6 &= 22\ 4x &= 28\ x &= 7ewline\text{The solution is } x = 7\text{, which is a valid length.} \end{aligned}\)

Key Concepts

Solving Equations

One of the core skills in geometry problem-solving is solving equations. These equations help us find unknown lengths, such as the length of a rectangle. When solving for a given variable, follow these steps:
1. Combine like terms. This makes the equation simpler.
2. Isolate the variable. You want to get the variable alone on one side of the equation.
These two principles are used in all four equations in the exercise. Here's a quick breakdown:

• For Equation A: We combined like terms:
  \(2x + 2(x-1) = 14 \rightarrow 2x + 2x - 2 = 14 \rightarrow 4x - 2 = 14\) then isolated the variable to find \(x = 4\).
• For Equation B: By rearranging and simplifying, we found that \(x = -\frac{17}{9}\). Since this is negative, it's not valid for a length.
• For Equation C: It simplified to \(5(x+2) + 5x = 10 \rightarrow 10x + 10 = 10 \rightarrow 10x = 0 \rightarrow x = 0\). Length can't be zero, making this invalid.
• For Equation D: We found \(2x + 2(x-3) = 22 \rightarrow 4x - 6 = 22 \rightarrow 4x = 28 \rightarrow x = 7\).

Total time dedicated to practicing these equations will sharpen your problem-solving skills!

Geometry Applications

Geometry is everywhere in our daily lives. We often solve geometric problems using algebraic equations. Understanding these connections can make learning more interesting.
In the given exercise, equations represent the perimeter or other geometric properties of a rectangle. Let's break down the applications:
- Perimeter of a rectangle is calculated as \(P = 2l + 2w\), where l is the length and w is the width. This relationship was evident in trying to equate total sums to a given perimeter.
- When solving real-life geometry applications, you’ll encounter similar steps:
1. Identify the geometric property you need to find.
2. Write an equation that represents that property.
3. Solve the equation to find your answer.
For instance, in equations A and D, we calculated lengths that matched a provided perimeter, ensuring these lengths were valid.

Length of a Rectangle

Determining the length of a rectangle is a straightforward process when you understand the basics of algebra and geometry. In these problems, we focus on:
- Validating the solutions: Any negative values or zero are not acceptable for geometric lengths.
- Real-life contexts: When working through these problems, imagine a physical object.
In the exercise:
- Equation A led us to \(x = 4\), which is a valid length.
- Equation B produced \(x = -\frac{17}{9}\). Since lengths can't be negative, this was invalid.
- Equation C resulted in \(x = 0\), which again is invalid because objects can't have zero length.
- Lastly, equation D showed \(x = 7\), another valid measurement.
Being attentive to the possible values of lengths and practicing through various problems helps build a strong foundation in both algebra and geometry.