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(a) complete the table of solutions. (b) graph the equation. $$ \begin{aligned} &y=-\frac{5}{6} x+8\\\ &\begin{array}{|r|r|} \hline \multicolumn{1}{|c|}{x} & y \\ \hline-6 & \\ \hline 0 & \\ \hline 6 & \\ \hline \end{array} \end{aligned} $$

Short Answer

Expert verified
For x = -6: y = 13, for x = 0: y = 8, for x = 6: y = 3. Plot the points and draw a line.

Step by step solution

01

Understand the equation

The given equation is \[ y = -\frac{5}{6}x + 8 \]. This is a linear equation in the form of \[ y = mx + b \], where \[ m = -\frac{5}{6} \] is the slope and \[ b = 8 \] is the y-intercept.
02

Substitute x = -6 into the equation

Calculate y by substituting \[ x = -6 \] into the equation:\[ y = -\frac{5}{6}(-6) + 8 \]\[ y = 5 + 8 \]\[ y = 13 \]
03

Substitute x = 0 into the equation

Calculate y by substituting \[ x = 0 \] into the equation:\[ y = -\frac{5}{6}(0) + 8 \]\[ y = 8 \]
04

Substitute x = 6 into the equation

Calculate y by substituting \[ x = 6 \] into the equation:\[ y = -\frac{5}{6}(6) + 8 \]\[ y = -5 + 8 \]\[ y = 3 \]
05

Complete the table

Fill in the calculated y values for the respective x values:\[ \begin{array}{|r|r|} \hline \multicolumn{1}{|c|}{x} & y \ \hline -6 & 13 \ \hline 0 & 8 \ \hline 6 & 3 \ \hline \end{array} \]
06

Graph the equation

To graph the equation \[ y = -\frac{5}{6}x + 8 \], plot the points (-6, 13), (0, 8), and (6, 3) on a coordinate plane. Draw a straight line passing through these points to complete the graph.

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Key Concepts

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

Table of Solutions
When dealing with a linear equation like \( y = -\frac{5}{6}x + 8 \), completing a table of solutions helps us understand how the value of \( y \) changes with different \( x \) values. For this equation, we calculated \( y \) values for \ x = -6 \, \ x = 0 \, and \ x = 6 \. We achieved this by substituting each \ x \ value into the equation and solving for \ y \. Here's a summary of the process:
  • When \ x = -6 \, \ y = 13 \.
  • When \ x = 0 \, \ y = 8 \.
  • When \ x = 6 \, \ y = 3 \.
By completing this table, we get pairs of \( x, y \) values that ultimately help us graph the equation. The points we can plot are \( (-6, 13) \, (0, 8) \, and \ (6, 3) \). This approach makes the graphing process much more straightforward and verifies that the line passes through each point correctly.
Graphing Linear Equations
Graphing a linear equation involves plotting points on a coordinate plane and drawing a line through those points. Using our equation \( y = -\frac{5}{6}x + 8 \), and the points from our table of solutions, we:
  • Plot the point \ (-6, 13) \ on the coordinate plane.
  • Plot the point \ (0, 8) \ on the coordinate plane.
  • Plot the point \ (6, 3) \ on the coordinate plane.
After plotting these points, draw a straight line through each one. Linear equations always form a straight line, regardless of the values of \( x \) and \( y \). The points confirm our equation is correctly graphed when the line passes through them perfectly. This method helps visually confirm the equation's solutions and better understand the relationship between \( x \) and \( y \).
Slope-Intercept Form
The equation \( y = -\frac{5}{6}x + 8 \) is in slope-intercept form, a common way to express linear equations. In slope-intercept form, an equation appears as \ y = mx + b \, where:
  • \ m \ is the slope of the line, which represents the rate of change. In our equation, \ m = -\frac{5}{6} \, indicating the line slopes downward at a ratio of \ -5 \ to \ 6 \.
  • \ b \ is the y-intercept, the point where the line crosses the y-axis. For our equation, \ b = 8 \ means the line crosses the y-axis at \ y = 8 \.
The slope tells us the steepness and direction of the line. A negative slope means the line decreases as \ x \ increases, moving from left to right. The y-intercept is helpful for starting the graph, offering a point where \ x = 0 \ and then use the slope to find other points. Understanding both components is crucial for graphing and interpreting linear equations, providing a visual way to understand algebraic relationships.

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