Problem 39
Question
The variables x and y vary directly. When x = 4, y = 24. Which equation correctly relates x and y? $$F. x=4 y$$ $$G.y=4 x$$ $$H.x=6 y$$ $$J.y=6 x$$
Step-by-Step Solution
Verified Answer
The equation that correctly relates x and y is J. \(y=6x\).
1Step 1: Understanding Direct Variation and the Given Condition
Direct variation is when two variables increase or decrease in sync with each other. Here, x and y are given to vary directly which means there is an equation in the form of \(y=kx\) where \(k\) is the constant of variation. Given that when \(x=4\) then \(y=24\), the aim is to find this constant \(k\).
2Step 2: Finding the Constant of Variation
Substitute \(x=4\) and \(y=24\) into the direct variation equation \(y=kx\). Thus, we have \(24=k*4\). Solving for \(k\), we get \(k = 24/4 = 6\). Thus, the direct variation equation in our case is \(y=6x\).
3Step 3: Comparing Against the Provided Options
The derived equation \(y=6x\) should be compared with the given four equations to find a match. The equation \(y=6x\) corresponds to option J. Thus, option J is the correct choice.
Key Concepts
Understanding the Constant of VariationExploring Variables in Direct VariationDefining Linear Equations through Direct Variation
Understanding the Constant of Variation
When working with direct variation, it's essential to understand the constant of variation. This constant, often represented by the symbol \( k \), unites the two variables in a direct relationship. It acts like a reliable bridge between them, determining their consistent rate of change.
In our exercise, since \( y = kx \), finding \( k \) allows us to express \( y \) in terms of \( x \). For example:
\( 24 = k \cdot 4 \). Solving this gives \( k = 6 \). This result tells us that with each increase in \( x \) by 1, \( y \) increases by 6, illustrating their direct connection in this relationship.
In our exercise, since \( y = kx \), finding \( k \) allows us to express \( y \) in terms of \( x \). For example:
- If \( x \) doubles, \( y \) also doubles, maintaining the ratio \( \frac{y}{x} = k \).
- The constant \( k \) ensures that for every unit increase in \( x \), \( y \) changes by \( k \) units.
\( 24 = k \cdot 4 \). Solving this gives \( k = 6 \). This result tells us that with each increase in \( x \) by 1, \( y \) increases by 6, illustrating their direct connection in this relationship.
Exploring Variables in Direct Variation
In direct variation, variables are the dynamic symbols that change value based on their relationships, embodying the concept of change. These variables, typically noted as \( x \) and \( y \), are core elements that allow us to describe and predict shifting patterns through equations.
In the context of our exercise, \( x \) and \( y \) were the variables. Here’s a closer look:
Understanding how these variables interact helps us capture the essence of direct variation.
In the context of our exercise, \( x \) and \( y \) were the variables. Here’s a closer look:
- \( x \): This is often considered the independent variable. It can change freely, determining how \( y \) will respond.
- \( y \): This is the dependent variable in our relationship. Its value is directly influenced by the value of \( x \).
Understanding how these variables interact helps us capture the essence of direct variation.
Defining Linear Equations through Direct Variation
Linear equations are fundamental expressions in math that describe a straight line when plotted on a graph. These equations take the form \( y = mx + c \), where \( m \) stands for slope, and \( c \) is the y-intercept. However, in direct variation involving two variables, our linear equation simplifies to \( y = kx \) since \( c = 0 \) (there's no y-intercept).
Within the exercise context, the derived equation \( y = 6x \) is a special kind of linear equation showing direct variation. Here’s why:
Within the exercise context, the derived equation \( y = 6x \) is a special kind of linear equation showing direct variation. Here’s why:
- The slope, denoted by \( k \), is constant, maintaining the direct relationship between \( y \) and \( x \).
- Graphically, \( y = 6x \) would result in a line passing through the origin, demonstrating its pure linearity and direct variation.
Other exercises in this chapter
Problem 39
Graph the function. $$ f(x)=4 x+1 $$
View solution Problem 39
Find the \(x\) -intercepts and the \(y\) -intercepts of the line. Graph the equation. Label the points where the line crosses the axes. $$ y=x+3 $$
View solution Problem 39
Write the equation in slope-intercept form. Then graph the equation. $$x-y+4=0$$
View solution Problem 39
Evaluate the expression. (Lesson 1.3) $$ 4 \cdot 5-2 \cdot 6 $$
View solution