Problem 15 - Algebra and Trigonometry

Question

13–22 ? Express the statement as an equation. Use the given information to find the constant of proportionality. \(M\) varies directly as \(x\) and inversely as \(y .\) If \(x=2\) and \(y=6,\) then \(M=5 .\)

Step-by-Step Solution

Answer

The constant of proportionality is 15.

Step 1: Understand Direct and Inverse Variation

When a quantity varies directly as another, it can be written as a multiplication (\(M= k \cdot x\) for some constant \(k\)). When a quantity varies inversely as another, it can be written as a division. If M varies directly with \(x\) and inversely with \(y\), the relationship is \(M = k \cdot \frac{x}{y}\).

Step 2: Identify the Given Values and Equation

The problem states that \(x=2\), \(y=6\), and \(M=5\). Substituting these values into the equation \(M = k \cdot \frac{x}{y}\), we get \(5 = k \cdot \frac{2}{6}\).

Step 3: Solve for the Constant of Proportionality \(k\)

To find \(k\), solve the equation from Step 2: \(5 = k \cdot \frac{2}{6}\). This simplifies to \(5 = \frac{2k}{6}\). You can simplify \(\frac{2}{6}\) to \(\frac{1}{3}\), yielding \(5 = \frac{k}{3}\).

Step 4: Solve the Equation for \(k\)

Multiply both sides of the equation \(5 = \frac{k}{3}\) by 3 to isolate \(k\). This gives \(k = 5 \cdot 3 = 15\).

Step 5: Verify the Expression

Now we have \(M = 15 \cdot \frac{x}{y}\). Substitute \(x=2\) and \(y=6\) back in to verify: \(M = 15 \cdot \frac{2}{6} = 15 \cdot \frac{1}{3} = 5\), which matches the given value of \(M\).

Key Concepts

Constant of Proportionality

In mathematics, the constant of proportionality plays a vital role when dealing with relationships between variables, especially in direct and inverse variations. Consider a scenario where two quantities are interconnected, such as in our exercise where the variable \(M\) varies directly with \(x\) and inversely with \(y\). This means the formula can be expressed as:

• \(M = k \cdot \frac{x}{y}\)

Here, \(k\) is the constant of proportionality. It stays consistent across changes in \(x\) and \(y\), allowing us to predict one variable based on the others if \(k\) is known. To find \(k\), values of other variables are substituted into the equation, as shown with \(x = 2, y = 6, \) and \(M = 5\). This results in the constant \(k = 15\), which acts as a scalar determining the strength of connection among the variables. This constant is crucial for defining the proportionality relationship and allows you to consistently apply the relationship formula for any other values of \(x\) and \(y\).

Algebraic Equations

Algebraic equations are mathematical statements that show the equality of two expressions, involving variables and constants. In this context, you can think of equations as tools that help us represent complex relationships in a simple and solvable form. In our exercise, the equation is:

• \(M = k \cdot \frac{x}{y}\)

This format indicates a mixed relationship: direct variation with \(x\) and inverse variation with \(y\). By substituting the known values of \(M\), \(x\), and \(y\), and isolating \(k\), we use algebraic manipulation to find the missing constant of proportionality. This is a reliable strategy to make sense of different variables' interconnections within mathematical problems.

Solving Equations

Solving equations involves finding the unknown variable or constant in your mathematical statement that makes the equation true. This process requires a sequence of steps, often involving manipulating the equation to isolate the unknown element so that its value can be determined. Let's illustrate this with our previous exercise.First, substitute the given values into the equation \(M = k \cdot \frac{x}{y}\). This becomes \(5 = k \cdot \frac{2}{6}\). Your goal is to solve for \(k\). Start by simplifying \(\frac{2}{6}\) to \(\frac{1}{3}\), converting the equation to \(5 = \frac{k}{3}\).

• Multiply both sides by 3: \(5 \times 3 = k\)
• This simplifies to \(k = 15\).

By following the steps of substitution, simplification, and isolation, you arrive at the solution: \(k = 15\). Remember, solving equations often involves a blend of arithmetic and logical reasoning to ensure that each manipulation is valid, ultimately leading to the correct solution.