Problem 36 - Algebra 2

Question

Find each constant of variation. Then find the value of \(y\) when \(x=-5\). $$ y=-36 \text { when } x=12 $$

Step-by-Step Solution

Answer

The constant of variation is \(k = -3\). The value of \(y\) when \(x = -5\) is 15.

Step 1: Identify the Direct Variation Equation

A direct variation between two variables can be represented by the equation \(y = kx\), where \(k\) is the constant of variation. Our goal is to find the value of \(k\) using the given values.

Step 2: Substitute Given Values to Find Constant of Variation

Substitute \(y = -36\) and \(x = 12\) into the direct variation equation to find \(k\). \[-36 = k \cdot 12\]Now, solve for \(k\) by dividing both sides of the equation by 12.\[k = \frac{-36}{12} = -3\]The constant of variation \(k\) is -3.

Step 3: Find the Value of y when x = -5

Use the constant of variation \(k = -3\) and the direct variation equation \(y = kx\) to find \(y\) when \(x = -5\).\[y = -3 \cdot (-5)\]Multiply to get the value of \(y\).\[y = 15\]

Key Concepts

Direct Variation

Understanding the concept of direct variation is fundamental in algebra. It describes a relationship between two variables in which one is a constant multiple of the other. Essentially, when one variable increases, the other increases by a proportional amount, and vice-versa. The general form of direct variation is represented by the equation \(y = kx\), where \(y\) and \(x\) are the variables in direct variation, and \(k\) is the constant of variation or the constant of proportionality.

For any given pair of numbers, if they exhibit direct variation, you can always find the constant \(k\), which remains the same throughout the relationship. Once you have established that two quantities vary directly, any changes in one quantity will predictably affect the other based on this constant multiplication factor. This consistency allows us to make predictions and solve problems involving direct variation.

Algebraic Equations

Algebraic equations are the backbone of algebra and serve as the foundation for expressing mathematical relationships. An equation is essentially a statement that two expressions are equal, with one or more variables representing unknown quantities. The beauty of algebraic equations lies in their versatility – they can describe everything from simple interactions like the cost of apples per pound to the complex relationships between physical forces.

When dealing with an algebraic equation, such as \(y = kx\), your main task is to isolate the variable of interest and solve for it. The process often involves performing the same mathematical operation on both sides of the equation to maintain equality. Clear understanding and manipulation of algebraic equations are crucial when trying to find unknown variables, like the constant of variation in direct variation problems.

Solving for Variables

Solving for variables is a critical skill in algebra. It involves manipulating an equation in order to determine the value of an unknown quantity. When we solve for a variable, we are rearranging the equation so that the variable is by itself on one side, and everything else is on the other side. This process typically uses operations such as addition, subtraction, multiplication, division, and even more complex techniques like factoring or using the quadratic formula, depending on the type of equation.

For instance, in the direct variation equation \(y = kx\), if you are given the values for \(y\) and \(x\), you can solve for the variable \(k\) by dividing both sides by \(x\). This action essentially 'undoes' the multiplication by \(x\), leaving the variable \(k\) isolated. Understanding how to correctly solve for variables is crucial not only for dealing with direct variation problems but for solving all kinds of algebraic equations.