Problem 65
Question
For each function, \(y\) varies directly as \(x\) If \(y=4\) when \(x=7,\) find \(y\) when \(x=-14\)
Step-by-Step Solution
Verified Answer
The value of y when x=-14 is -8.
1Step 1: Identify the Direct Variation Relationship
First, understand that 'y varies directly as x' means y can be expressed as y = kx, where k is the constant of variation.
2Step 2: Determine the Constant of Variation
Use the initial condition y=4 when x=7 to find the constant of variation (k). Set up the equation 4 = k(7). Solve for k by dividing both sides of the equation by 7, which gives k = 4/7.
3Step 3: Apply the Constant of Variation
Now that you have found k to be 4/7, use this constant to find y when x=-14, by substituting k into the direct variation equation to get y = (4/7)(-14).
4Step 4: Calculate the New Value of y
Multiply the constant k by the new value of x to find y. So, y = (4/7)(-14) simplifies to y = -8.
Key Concepts
Constant of VariationProportional RelationshipsAlgebraic Equations
Constant of Variation
When we talk about direct variation, we are referring to a simple relationship where one variable changes by a constant amount as another variable changes. The constant of variation, denoted as 'k', is vital in understanding this relationship.
Think of it as a consistent ratio that relates values of two variables. In a direct variation scenario, when the value of one variable is multiplied by the constant of variation, the result is the value of the other variable. For instance, in our exercise, the equation for direct variation was expressed as \(y = kx\), where 'k' is the constant of variation.
To nail down 'k', you just need to know one set of values for 'x' and 'y'. As seen in our exercise, when \(x = 7\) and \(y = 4\), we arranged these numbers into the equation to solve for 'k', ending up with \(k = \frac{4}{7}\). Once we have 'k', we can predict 'y' for any given 'x', maintaining this set ratio.
Think of it as a consistent ratio that relates values of two variables. In a direct variation scenario, when the value of one variable is multiplied by the constant of variation, the result is the value of the other variable. For instance, in our exercise, the equation for direct variation was expressed as \(y = kx\), where 'k' is the constant of variation.
To nail down 'k', you just need to know one set of values for 'x' and 'y'. As seen in our exercise, when \(x = 7\) and \(y = 4\), we arranged these numbers into the equation to solve for 'k', ending up with \(k = \frac{4}{7}\). Once we have 'k', we can predict 'y' for any given 'x', maintaining this set ratio.
Proportional Relationships
A proportional relationship is a key concept that directly relates to the idea of direct variation. It’s when two quantities always maintain the same ratio, or in other words, their relationship can be described by a constant of proportionality. This constant is the 'k' we talked about before.
In real-world terms, if you're driving at a steady speed, the distance you travel is directly proportional to the time you've been driving. Here, your speed would be the constant of proportionality.
In the direct variation equation \(y = kx\), 'x' and 'y' maintain a consistent ratio defined by the constant 'k'. This ensures that for every unit increase in 'x', 'y' increases by 'k' units. This proportionality makes it incredibly straightforward to track how one variable affects another and allows for easy predictions and calculations as shown in our example, which is especially useful in fields like physics, economics, and even in everyday problem solving.
In real-world terms, if you're driving at a steady speed, the distance you travel is directly proportional to the time you've been driving. Here, your speed would be the constant of proportionality.
In the direct variation equation \(y = kx\), 'x' and 'y' maintain a consistent ratio defined by the constant 'k'. This ensures that for every unit increase in 'x', 'y' increases by 'k' units. This proportionality makes it incredibly straightforward to track how one variable affects another and allows for easy predictions and calculations as shown in our example, which is especially useful in fields like physics, economics, and even in everyday problem solving.
Algebraic Equations
Algebraic equations are the foundation of solving problems in algebra. They consist of variables and constants, and they define the relationships between these quantities. The process of solving algebraic equations involves finding the value of the unknowns that make the equation true.
In the context of direct variation, the algebraic equation we examine is \(y = kx\), where 'k' is the constant of variation. Once we determine 'k', the equation serves as a model that describes how 'y' is influenced by any value of 'x'.
When working through our exercise, we initially used known values of 'x' and 'y' to solve for 'k', and then we applied 'k' to a new value of 'x' to find 'y'. This is a straightforward example of using algebraic equations to understand and describe relationships between variables. It shows how algebra can be used to decipher patterns and make predictions.
In the context of direct variation, the algebraic equation we examine is \(y = kx\), where 'k' is the constant of variation. Once we determine 'k', the equation serves as a model that describes how 'y' is influenced by any value of 'x'.
When working through our exercise, we initially used known values of 'x' and 'y' to solve for 'k', and then we applied 'k' to a new value of 'x' to find 'y'. This is a straightforward example of using algebraic equations to understand and describe relationships between variables. It shows how algebra can be used to decipher patterns and make predictions.
Other exercises in this chapter
Problem 64
Graph each inequality on a coordinate plane. $$ x
View solution Problem 65
Solve each inequality. Graph the solution on a number line. $$ 6-2 x>2 $$
View solution Problem 65
Find the value of \(a\) that makes each system a dependent system. \(\left\\{\begin{array}{l}{3 y=2 x} \\ {6 y-a-4 x=0}\end{array}\right.\)
View solution Problem 65
Write an equation for each line. $$ m=-\frac{2}{3} ; \text { contains }(-9,4) $$
View solution