Problem 7 - Technical Mathematics with Calculus

Question

A student sold used skis and boots for \(\$ 210,\) getting 4 times as much for the boots as for the skis. What was the price of each?

Step-by-Step Solution

Answer

The skis were sold for \$42 and the boots for \$168.

Step 1: Define Variables

Let's call the price of the skis \(s\) and the price of the boots \(b\). The problem tells us that the student got 4 times as much for the boots as for the skis, so we can write the equation \(b = 4s\).

Step 2: Write the Total Price Equation

We also know the total price for both skis and boots is \$210. We can write this as another equation: \(s + b = 210\).

Step 3: Substitute the Value of \(b\) in the Total Price Equation

Substitute \(4s\) for \(b\) in the total price equation to get a single variable equation: \(s + 4s = 210\).

Step 4: Solve the Equation

Combine like terms and solve for \(s\): \(5s = 210\). Divide both sides by 5 to find \(s\): \(s = \frac{210}{5} = 42\).

Step 5: Find the Price of the Boots

Now that we know \(s\), we can find \(b\): \(b = 4s = 4 \times 42 = 168\).

Key Concepts

Algebraic Equations

Algebraic equations are the backbone of solving mathematical problems that involve finding unknown values. They consist of algebraic expressions separated by an equal sign, indicating that the expressions on both sides have the same value. In the context of our example, the equation used is \( b = 4s \) which represents the relationship between the price of boots (\( b \) and the price of skis (\( s \) such that the boots are four times as expensive as the skis.

The power of algebraic equations lies in their ability to create a model of real-world scenarios. By translating a word problem about selling skis and boots into a mathematical equation, one can use algebraic methods to find concrete numbers that solve the problem. The equation \( s + b = 210 \) condenses the information about the total sales into a workable mathematical form. To master algebraic equations, students should practice setting up equations that accurately represent word problems and then learn various techniques to solve them.

Variable Substitution

Variable substitution is a method used to simplify equations and solve for unknowns by replacing one variable with another expression. It's an essential skill in solving systems of equations, where two or more equations with two or more variables interact. In our example, we start with two equations: \( b = 4s \) and \( s + b = 210 \).

To solve these equations, we substitute the expression for \( b \) from the first equation into the second equation. This reduces the system to a single equation with one variable, \( s + 4s = 210 \) which simplifies the problem and makes it easier to solve. By understanding and applying variable substitution, students can reduce complex systems into simpler ones, making the path to the solution clearer. It is important to learn variable substitution, as it is widely used not just in algebra, but in various branches of mathematics and science.

Problem-Solving in Mathematics

Problem-solving in mathematics is a critical thinking process that involves understanding the problem, devising a plan, carrying out the plan, and then looking back to see if the solution makes sense. In our exercise, the problem is to find the price of each item - the skis and the boots. To approach this, one must define variables to represent unknown prices, establish equations based on the given information, and then use algebraic techniques to solve these equations.

Looking at the big picture, problem-solving skills are developed by practicing a wide variety of problems and learning multiple solving strategies like drawing diagrams, making lists, looking for patterns, and using algebraic equations. Improving one's problem-solving skills can lead to better mathematical intuition and the ability to tackle more complex mathematical concepts.