Problem 21 - Algebra 2 and Trigonometry

Question

In \(18-23,\) write and solve an equation or an inequality to solve the problem. At a parking garage, parking costs \(\$ 5\) for the first hour and \(\$ 3\) for each additional hour or part of an hour. Mr. Kanesha paid \(\$ 44\) for parking on Monday. For how many hours did Mr. Kanesha park his car?

Step-by-Step Solution

Answer

Mr. Kanesha parked for 14 hours.

Step 1: Set Up Equation

To find out how many hours Mr. Kanesha parked, we need to formulate an equation. The first hour costs \\(5 and each additional hour costs \\)3. Let \( h \) be the total number of hours Mr. Kanesha parked. Since he pays for the first hour separately, we consider it and then add the \( h-1 \) additional hours he parked at \\(3 each. The total cost equation is: \[ \text{Total Cost} = 5 + 3(h - 1) \] Given that the total cost is \\)44, we set up the equation: \[ 5 + 3(h - 1) = 44 \]

Step 2: Simplify the Equation

Simplify the equation by distributing the 3 into \((h - 1)\): \[ 5 + 3h - 3 = 44 \] Simplify further by combining like terms (\(5 - 3 = 2\)): \[ 3h + 2 = 44 \]

Step 3: Solve for h

To isolate \( h \), subtract 2 from both sides of the equation: \[ 3h = 44 - 2 \] \[ 3h = 42 \] Next, divide both sides by 3 to solve for \( h \): \[ h = \frac{42}{3} \] \[ h = 14 \]

Step 4: Conclusion

Mr. Kanesha parked his car for 14 hours in total.

Key Concepts

Linear Equations

Linear equations are equations where the highest exponent of the variable is one. They form a straight line when graphed and are typically in the format of \(ax + b = c\). In the context of our problem, linear equations help us translate a word problem into a mathematical expression that can be solved systematically.
Linear equations are used when dealing with problems that involve constants and variables with linear relationships. In Mr. Kanesha's parking situation, the equation \(5 + 3(h - 1) = 44\) helped to express how the total parking cost is calculated based on the number of hours.
Working through linear equations requires setting them up based on given information, simplifying the equation, and then solving for the unknown variable. The variable in this problem, \(h\), represents the total hours parked.

Substitution Method

The substitution method is a powerful tool in solving systems of equations, but it can also aid in single-variable equations by rephrasing or making use of known information. Here, instead of directly solving for \(h\), we set up an equation that naturally accounts for the cost breakdown.
This method allowed us to substitute the given values into the equation correctly. By understanding that the first hour is a fixed cost of \\(5 and each additional hour costs \\)3, we developed the expression \(5 + 3(h - 1)\). Here, substituting the known total cost (\$44) helps us articulate how the variable \(h\) works within this financial context.
Substitution simplifies the complexity of dealing with multiple changing variables, offering clarity by using known quantities and relationships directly in the equations.

Simplifying Expressions

Simplifying expressions involves breaking down a complex equation to its most concise form, making it easier to solve. This usually means combining like terms and distributing coefficients over terms within brackets.
In the exercise, the expression \(5 + 3(h - 1) = 44\) was simplified first by distributing the 3 to terms within the parentheses, giving \(3h - 3\), and then combining with the 5 to get \(3h + 2 = 44\).
Each simplification step reduces the complexity, making the equation easier to work with. It's crucial when simplifying to pay attention to signs and careful arithmetic to avoid mistakes. Simplification is a foundational skill in algebra that applies in equations, inequalities, and more, fostering clearer understanding and easier solution derivations.