Problem 13 - College Algebra

Question

Determine the constant of variation for each stated condition. \(y\) varies directly as \(x,\) and \(y=75\) when \(x=3\)

Step-by-Step Solution

Answer

The constant of variation \(k\) is 25.

Step 1: Substitute given values

Insert \(x = 3\) and \(y = 75\) into the equation \(y = kx\). So we get \(75 = 3k\).

Step 2: Solve for k

To find the value of \(k,\) divide both sides of the equation by 3. We get \(k = 75 / 3.\)

Step 3: Calculate k

After performing the division, we find that \(k = 25.\)

Key Concepts

Understanding Direct Variation

When we speak of direct variation, we're referring to a specific kind of relationship between two variables where one is a multiple of the other. In simpler terms, if one variable increases, the other increases at a consistent rate, and this rate is the constant of variation.

In algebra, we often use the equation \(y = kx\) to express this relationship, where \(y\) is the dependent variable, \(x\) is the independent variable, and \(k\) is the constant of variation. This constant can be any real number and basically tells you how much \(y\) changes when \(x\) changes.

Consider this real-life example: if a car travels at a constant speed, the distance covered (dependent variable) varies directly as the time traveled (independent variable). The car's speed would be the constant of variation.

In the textbook problem provided, we're asked to determine this constant when given a set of conditions. It's a fine example of seeing how direct proportionality works in algebraic terms.

Solving Algebraic Equations

Algebraic equations are the cornerstone of solving problems in algebra. These are mathematical statements that show the equality of two expressions with one or more unknowns, typically denoted by letters like \(x\) or \(y\) .

In order to find the value of these unknowns, we manipulate the equation using algebraic operations such as addition, subtraction, multiplication, and division while adhering to the rules of algebra.

The primary goal when dealing with an algebraic equation, like the one in our exercise \(y = kx\) , is to isolate the variable we're trying to solve for — in this case, \(k\) . We want to perform operations on both sides of the equation to maintain equality until we've successfully 'solved' for our unknown variable.

The right strategy depends on the form of the equation and the variables involved. In the direct variation problem from our textbook, solving for the constant of variation requires an understanding of these algebraic principles.

Solving for k, the Constant of Variation

The final step in our journey to understand direct variation is actually solving for the constant \(k\) . In the textbook example, we're given certain values for \(y\) and \(x\) and have to work out what \(k\) must be.

The process of solving for \(k\) is pretty straightforward if we follow the steps methodically. First, we substitute the provided values into our direct variation equation. As per the given solution, when we put \(x = 3\) and \(y = 75\) into \(y = kx\) , we get \(75 = 3k\) .

The next step is about using basic algebra to isolate \(k\) . Here, we divide both sides of the equation by 3, which gives us \(k = 75 / 3\) . After performing the division, you find that \(k = 25\) . This tells us for every single unit increase in \(x\) , \(y\) will increase by 25 units. This value of \(k\) is crucial because it's what defines the relationship between \(x\) and \(y\) under the rules of direct variation, and understanding how to solve for it is essential in various fields of math and science.