Problem 52
Question
The variables x and y vary directly. Use the given values to write an equation that relates x and y. $$x=7, y=5$$
Step-by-Step Solution
Verified Answer
The equation that relates \(x\) and \(y\) is \(y = (5/7)x\).
1Step 1: Understanding Direct Variation
In a direct variation, as one quantity increases, the other quantity increases at a constant rate. The formula for a direct variation is \(y = kx\), where \(k\) is the constant of variation, \(y\) is the dependent variable and \(x\) is the independent variable.
2Step 2: Find the Constant
With given equations \(x = 7\) and \(y = 5\), substitute these values into \(y = kx\). It becomes \(5 = k * 7\). Solve for \(k\) by dividing both sides of the equation by 7. You get \(k = 5/7\).
3Step 3: Write the variation
Substitute \(k = 5/7\) back into the direct variation formula \(y = kx\). This results in the equation that explains the direct variation between \(x\) and \(y\), which is \(y = (5/7)x\).
Key Concepts
Constant of VariationDependent and Independent VariablesWriting Direct Variation Equations
Constant of Variation
In the context of direct variation, the constant of variation is a key concept that indicates the fixed rate at which two quantities change together. Whenever you deal with a scenario where one variable increases or decreases in direct proportion to another, you're looking at a case of direct variation.
Mathematically, this relationship is represented as \(y = kx\), where \(y\) and \(x\) are the variables in direct variation, and \(k\) is the constant of variation. It is this 'k' that we need to find to understand how the variables are connected.
In the example where \(x = 7\) and \(y = 5\), we can find the constant of variation by rearranging the formula to solve for 'k': \( k = \frac{y}{x} = \frac{5}{7} \). This constant \(\frac{5}{7}\) remains the same no matter what values of \(x\) and \(y\) you have, as long as they maintain the direct variation relationship.
Mathematically, this relationship is represented as \(y = kx\), where \(y\) and \(x\) are the variables in direct variation, and \(k\) is the constant of variation. It is this 'k' that we need to find to understand how the variables are connected.
In the example where \(x = 7\) and \(y = 5\), we can find the constant of variation by rearranging the formula to solve for 'k': \( k = \frac{y}{x} = \frac{5}{7} \). This constant \(\frac{5}{7}\) remains the same no matter what values of \(x\) and \(y\) you have, as long as they maintain the direct variation relationship.
Dependent and Independent Variables
When exploring direct variation, it's crucial to distinguish between dependent and independent variables. In a direct variation relationship, the independent variable is the one you can manipulate or choose freely, while the dependent variable is the one that changes in response to the independent variable.
For instance, in the equation \(y = kx\), \(x\) is considered the independent variable, and \(y\) is the dependent variable. This means that as you change the value of \(x\) (the independent variable), the value of \(y\) (the dependent variable) alters according to the constant of variation 'k'. The role of each variable is crucial for effectively writing equations and for comprehending how changes in one variable affect the other.
For instance, in the equation \(y = kx\), \(x\) is considered the independent variable, and \(y\) is the dependent variable. This means that as you change the value of \(x\) (the independent variable), the value of \(y\) (the dependent variable) alters according to the constant of variation 'k'. The role of each variable is crucial for effectively writing equations and for comprehending how changes in one variable affect the other.
Writing Direct Variation Equations
When it comes to writing direct variation equations, the process involves identifying the constant of variation and knowing which variable is dependent on the other. If you have a pair of values, like in our example with \(x = 7\) and \(y = 5\), you first determine the constant by dividing \(y\) by \(x\).
Once the constant \(k\) is found, \(k = \frac{5}{7}\), you express the relationship with an equation: \(y = kx\). Substituting our specific 'k' back into this formula gives us \(y = \frac{5}{7}x\), which is the equation for the direct variation. It can now be used to predict the value of the dependent variable (y) for any given value of the independent variable (x).
This approach isn't just about plugging numbers into an equation; it is about understanding and expressing the consistent relationships between variables.
Once the constant \(k\) is found, \(k = \frac{5}{7}\), you express the relationship with an equation: \(y = kx\). Substituting our specific 'k' back into this formula gives us \(y = \frac{5}{7}x\), which is the equation for the direct variation. It can now be used to predict the value of the dependent variable (y) for any given value of the independent variable (x).
This approach isn't just about plugging numbers into an equation; it is about understanding and expressing the consistent relationships between variables.
Other exercises in this chapter
Problem 52
Write the quadratic equation in standard form. Solve using the quadratic formula. $$-16 b=-8 b^{2}-8$$
View solution Problem 52
The consumption of Swiss cheese in the United States from 1970 to 1996 can be modeled by \(P=-0.002 t^{2}+0.056 t+0.889\) where \(P\) is the number of pounds pe
View solution Problem 52
SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=3 x^{2}-2 x $$
View solution Problem 53
Use a calculator to evaluate the expression. Round the results to the nearest hundredth. $$\frac{7 \pm 0.3 \sqrt{12}}{-6}$$
View solution