Problem 22 - Algebra 1: Concepts and Skills

Question

In Exercises 16–24, the variables x and y vary directly. Use the given values to write an equation that relates x and y. $$ x=-1, y=-1 $$

Step-by-Step Solution

Answer

The equation of direct variation that describes the relationship between x and y is y=x.

Step 1: Apply the direct variation formula

Let's apply the formula for direct variation, y = kx, and substitute the given x and y values into the formula accordingly. This leads to -1 = k*(-1).

Step 2: Solve for k

By solving this simple equation, you get that k equals to 1. This is done by dividing both sides of the equation by -1.

Step 3: Substitute k into the direct variation formula

Now that we have determined the value for k, we can substitute it back into the equation for direct variation. This gives us the equation y = x, which describes the direct relationship between x and y.

Key Concepts

Direct Variation Formula

Understanding direct variation is crucial when you're dealing with proportional relationships in algebra. It is the concept that describes a very straightforward relationship between two variables: when one variable increases or decreases, the other does so as well, at a constant rate. The formula that represents direct variation is a simple equation \( y = kx \), where \( y \) and \( x \) are the variables that are directly proportional to each other, and \( k \) is the constant of variation or the proportionality constant. This constant remains the same throughout the relationship.

For example, if we state that \( y \) varies directly as \( x \) and we have a pair of values for \( y \) and \( x \) such as \( x = -1 \) and \( y = -1 \) from the exercise, we can plug these into the formula to find our constant \( k \). After substitution, our equation looks like \( -1 = k*(-1) \). Solving for \( k \) gives us a constant of 1, meaning for every unit that \( x \) increases or decreases, \( y \) will do the same proportionally.

Algebraic Equations

Algebraic equations are mathematical statements that show the equality between two expressions. They are composed of variables, coefficients, and constants, and can range from the simple, such as linear equations, to the more complex including quadratic or exponential equations. In the context of direct variation and the exercise provided, we encounter a linear equation, which is fundamental to understanding algebraic relationships.

When solving for the variable, such as \( k \) in the direct variation formula, remember to perform the same operation on both sides of the equation to maintain the balance. If \( y = kx \) and we know the values of \( y \) and \( x \) we can solve for \( k \) by dividing both sides of the equation by the value of \( x \) (as long as \( x \) isn't zero, because division by zero is undefined). In the step by step solution we provided, dividing \( -1 \) by \( -1 \) leaves us with \( k = 1 \). The process of isolation is a vital skill in algebra.

Variable Relationships

In algebra, understanding the relationships between variables is essential. Direct variation is a type of variable relationship where two variables move in tandem—when one increases, the other increases; when one decreases, the other decreases. Another common relationship is inverse variation, where one variable increases while the other decreases.

Recognizing patterns in variable relationships can help you solve algebraic equations more effortlessly. For instance, if you know that \( y \) varies directly as \( x \) and you are given corresponding values, you can determine the exact equation that models their relationship. Moreover, identifying these patterns allows you to make predictions and understand the behavior of variables within a system. It's not just about knowing that \( y = kx \) for direct variation, but also about grasping what this means for the change in \( y \) when \( x \) changes, which is a critical aspect of problem-solving in mathematics.