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Chapter 1: Problem 24

For Exercises 20鈥�28, answer Parts a and b. a. What is the constant difference between the \(y\) values as the \(x\) values increase by 1\(?\) b. What is the constant difference between the \(y\) values as the \(x\) values decrease by 2\(?\) $$ y=5 x $$

Short Answer

Expert verified
The constant difference for increasing x by 1 is 5. The constant difference for decreasing x by 2 is -10.

Step by step solution

01

- Identify the Equation

The given equation is: \[ y = 5x \]. This is a linear equation where the slope is the coefficient of \( x \), which is 5.
02

- Determine the Constant Difference for Increasing x by 1

When \( x \) increases by 1, the change in \( y \) can be found by substituting the new and old values of \( x \) into the equation. Let \( x_1 \) be the original value and \( x_2 = x_1 + 1 \) be the new value. Then, \[ y_2 = 5(x_1 + 1) \] and \[ y_1 = 5x_1 \]. The difference, \( \text{Diff}_1 \), is: \[ \text{Diff}_1 = y_2 - y_1 = 5(x_1 + 1) - 5x_1 = 5x_1 + 5 - 5x_1 = 5 \]. Therefore, the constant difference for increasing \( x \) by 1 is 5.
03

- Determine the Constant Difference for Decreasing x by 2

When \( x \) decreases by 2, the change in \( y \) can be found similarly. Let \( x_1 \) be the original value and \( x_2 = x_1 - 2 \) be the new value. Then, \[ y_2 = 5(x_1 - 2) \] and \[ y_1 = 5x_1 \]. The difference, \( \text{Diff}_2 \), is: \[ \text{Diff}_2 = y_2 - y_1 = 5(x_1 - 2) - 5x_1 = 5x_1 - 10 - 5x_1 = -10 \]. Therefore, the constant difference for decreasing \( x \) by 2 is -10.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope
In the context of linear equations, the slope is an important concept. It indicates how steep the line is and the direction it is moving. Think of the slope as the rate at which one variable changes with respect to another. For the equation \( y = 5x \), the slope is \(5\). This value means that for every increase of 1 unit in \(x\), \(y\) increases by 5 units. The slope tells you how fast the \(y\) values are changing, all in a direct linear relationship with \(x\).

In simpler terms:
  • If the slope is positive, like in this example (\(m = 5\)), the line goes upwards as you move from left to right.
  • If the slope were negative, the line would go downwards.
  • If the slope is zero, the line is horizontal.
Constant Difference
One key feature of linear equations is the constant difference between \(y\)-values. This concept is evident when you adjust \(x\) by a known amount. For the given equation, \( y = 5x \), we calculated the difference in \(y\)-values when \(x\) changes.

For an increase in \(x\) by 1, the change in \(y\) is 5. This is because the slope \( m \) is 5. So, the constant difference will always be 5 as long as \(x\) increments by 1.

For a decrease in \(x\) by 2, you find the change like so:
  • Original \( y = 5x \)
  • New \( y_2 = 5(x_1 - 2)\)
  • Difference, \( \text{Diff}_2 = -10 \)
The constant difference for a 2 unit drop in \(x\) is therefore -10. The negative sign shows the direction of change.
Variable Change
Understanding how changing \(x\) impacts \(y\) is crucial in mastering linear equations. In our example, the equation \( y = 5x \) directly ties changes in \(x\) to changes in \(y\). Every time you change \(x\), you can quickly determine the new \(y\) by using the equation.

Here's a simple breakdown:
  • If \(x\) increases by 1: Use \( y = 5(x + 1) \)
  • If \(x\) decreases by 2: Use \( y = 5(x - 2) \)
This method leads you to see how each unit change in \(x\) corresponds to a direct and linear change in \(y\), all governed by the slope of the equation. Linear equations provide a predictable, straightforward relationship between variables, simplifying the process of predicting changes.

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Most popular questions from this chapter

Give the slope-intercept form of each equation, and tell which of these eight equations describe the same relationship. a. \(4 x-2 y=4\) b. \(2 y-4 x=4\) c. \(2 x-y=2\) d. \(y-2 x=2\) e. \(y-2 x=-2\) f. \(y=2 x-2\) g. \(y=2 x+2\) h. \(4 x+2 y=4\)

Find an equation of the line passing through the given points. $$ (2,7) \text { and }(6,6) $$

You can tell whether a particular point might be on a line by graphing it and seeing whether it seems to lie on the line. But to know for certain whether a particular point is on a line鈥攁nd not just close to it鈥攜ou must test whether its coordinates satisfy the equation for that line. a. Graph the equation \(y=\frac{13}{8} x-3.\) b. Using the graph alone, decide which points below look like they might be on the line. You may want to plot the points. $$(0,-3) \quad(3,2) \quad(4,4) \quad(5,5) \quad(8,10) $$ c. For each point, substitute the coordinates into the equation and evaluate to determine whether the point satisfies the equation. Which points, if any, are on the line?

The lines for these three equations all pass through a common point. $$y=\frac{x}{2}-1 \quad y=-\frac{2 x}{3}+6 \quad y=-\frac{x}{6}+3$$ a. Draw graphs for the three equations, and find the common point. b. Verify that the point you found satisfies all three equations by substituting the x- and y-coordinates into each equation.

For Exercises 20鈥�28, answer Parts a and b. a. What is the constant difference between the \(y\) values as the \(x\) values increase by 1\(?\) b. What is the constant difference between the \(y\) values as the \(x\) values decrease by 2\(?\) $$ y=\frac{1}{2} x $$
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