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Chapter 2: Problem 64

Find the value of \(y\) if a linear function goes through the following points and has the following slope: \((10, y),(25,100), m=-5\)

Short Answer

Expert verified
The value of \( y \) is 175.

Step by step solution

01

Understand the Linear Equation

A linear function can be represented as \( y = mx + c \), where \( m \) is the slope, and \( c \) is the y-intercept. We are given the slope \( m = -5 \) and two points: \((10, y)\) and \((25, 100)\).
02

Use Point-Slope Form

The point-slope form of a line is given by \( y - y_1 = m(x - x_1) \). We can use the point \((25, 100)\) to apply this formula. So, \( y - 100 = -5(x - 25) \).
03

Simplify the Equation

Substitute the coordinates of the first point into the equation to find \( y \). First, simplify the expression: \( y - 100 = -5(x - 25) \) becomes \( y - 100 = -5x + 125 \).
04

Solve for Y

To find \( y \) when \( x = 10 \), substitute \( x = 10 \) into the equation: \( y = -5(10) + 225 \). This simplifies to \( y = -50 + 225 = 175 \).

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form of a linear equation is a way of expressing the equation of a line. In this form, the equation is written as:\( y = mx + c \).Understanding this form is crucial for working with linear equations. Here:
  • \( m \) is the slope of the line, representing how steep the line is. In simpler terms, it tells us how much \( y \) changes for a unit change in \( x \).
  • \( c \) is the y-intercept, the point where the line crosses the y-axis. When \( x = 0 \), \( y \) is equal to \( c \).
The slope-intercept form is particularly useful because it immediately gives you the slope and y-intercept, making it easy to graph the equation and understand the line's behavior.
In the original exercise, once the line's equation is simplified, it can be compared to this form to easily identify the characteristics of that specific linear function.
Point-Slope Form
The point-slope form is another convenient way to express the equation of a line, especially when you know a point on the line and the slope. The equation takes the form of:\( y - y_1 = m(x - x_1) \).Here, \((x_1, y_1)\) is a known point on the line, and \( m \) is the slope. This form directly ties in a specific point, allowing you to form the line's equation even if the y-intercept is unknown.
  • A major benefit of the point-slope form is that it simplifies when you have a specific point and a slope, allowing for easy substitution into the equation.
  • This form is ideal for initial computation before transitioning to the slope-intercept form for further interpretation.
In the given problem, the point-slope form was utilized with the point \((25, 100)\) to find the equation of the line. It helped pinpoint the line equation, which is crucial for determining other unknown values, such as the missing \( y \) at \( x = 10 \).
Solving for a Variable
Solving for a variable means isolating it on one side of the equation to find its value. It's a basic yet essential skill in algebra that has wide applications.In the exercise, we had to solve for \( y \) when the line passed through specific points and had a defined slope.
  • Start by substituting known values into the equation derived from a particular form (like point-slope or slope-intercept).
  • Rearrange the equation to make the variable of interest the subject of the formula.
  • Perform operations step-by-step to isolate the variable. These might include addition, subtraction, multiplication, division, or a combination of those.
Specifically, for this problem, once the equation was formed, by substituting \( x = 10 \), the operations were performed to find \( y \). Each algebraic manipulation brought us closer to isolating \( y \) and determining its value within the context of the equation. This process bridges numerical value with theoretical understanding, as it allows us to find specific values for specific instances.

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Most popular questions from this chapter

For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: $$-2 x+y=3$$ $$3 x+\frac{3}{2} y=5$$

For the two linear functions, find the point of intersection: \(x=y+2\) \(2 x-3 y=-1\)

For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population, year) for specific recorded years: $$(2500,2000),(2650,2001),(3000,2003),(3500,2006),(4200,2010)$$ Predict when the population will hit 8,000.

For the following exercises, use each set of data to calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to 3 decimal places of accuracy. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {100} & {80} & {60} & {55} & {40} & {20} \\ \hline y & {2000} & {1798} & {1589} & {1580} & {1390} & {1202} \\\ \hline\end{array}$$

Determine whether the algebraic equation is linear. \(2 x+3 y=7\)
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