Part A: Graph the system of linear equations. Part B: Use the graph created in Part A to determine the solution to the system. Part C: Algebraically verify the solution from a Part B
x + 6y = 6 y = 1/3x - 2
To graph the system of linear equations, we need to plot the lines represented by each equation on the same coordinate plane.
Part A:
The first equation is x + 6y = 6. To plot this equation, we can rewrite it in slope-intercept form (y = mx + b) by isolating y:
6y = -x + 6
y = (-1/6)x + 1
Now, we can determine two points on this line. We can choose any values for x and calculate the corresponding y values using the equation. Let's choose two easy values for x, such as x = 0 and x = 6:
For x = 0:
y = (-1/6)(0) + 1
y = 1
So we have the point (0, 1) on the line.
For x = 6:
y = (-1/6)(6) + 1
y = 0
So we have the point (6, 0) on the line.
Plotting these two points and drawing a line through them, we get the graph of the first equation.
The second equation is y = (1/3)x - 2, which is already in slope-intercept form. We can choose two more values for x and calculate y:
For x = 0:
y = (1/3)(0) - 2
y = -2
So we have the point (0, -2) on the line.
For x = 6:
y = (1/3)(6) - 2
y = 0
So we have the point (6, 0) on the line.
Plotting these two points and drawing a line through them, we get the graph of the second equation.
Part B:
To determine the solution to the system of equations using the graph, we need to find the point where the two lines intersect. This point represents the values of x and y that satisfy both equations simultaneously. In this case, it seems that the two lines intersect at the point (6, 0).
Part C:
To algebraically verify the solution found in Part B, we substitute the x and y values from the intersection point into both equations and check if they are true.
For the first equation, x + 6y = 6:
6 + 6(0) = 6
6 = 6
This equation is true.
For the second equation, y = (1/3)x - 2:
0 = (1/3)(6) - 2
0 = 2 - 2
0 = 0
This equation is also true.
Therefore, the solution (x, y) = (6, 0) correctly satisfies both equations algebraically, confirming the solution found using the graph in Part B.
To graph the system of linear equations, we'll start with Part A:
Part A: Graphing the system of linear equations.
1. Let's start with the first equation: x + 6y = 6.
To make it easier to graph, let's rearrange the equation to solve for y:
6y = 6 - x
y = (6 - x) / 6
Plotting this equation on a coordinate plane, we get a line.
Choose a few values for x and calculate the corresponding y:
For x = 0, y = (6 - 0) / 6 = 6/6 = 1
For x = 6, y = (6 - 6) / 6 = 0/6 = 0
Plot these points (0, 1) and (6, 0) and draw a straight line passing through them.
2. Now let's move on to the second equation: y = 1/3x - 2.
This equation is already in slope-intercept form (y = mx + b), where the coefficient of x is the slope and b is the y-intercept.
From this equation, we can see that the slope is 1/3, and the y-intercept is -2.
Therefore, we can plot the y-intercept at (0, -2) and use the slope to find more points.
For example, if we go one unit to the right (x = 3), we go up 1/3 of a unit (y = -1 2/3).
Plot these points (0, -2) and (3, -1 2/3) and draw a straight line passing through them.
Now that we have graphed both equations, let's move on to Part B:
Part B: Determining the solution from the graph.
To solve the system of linear equations graphically, we look for the point of intersection on the graph. This point represents the solution.
Looking at the graph, we can see that the lines intersect at the point (3, -1).
So the solution to the system of linear equations is x = 3 and y = -1.
Finally, let's move on to Part C:
Part C: Algebraically verifying the solution from Part B.
We'll substitute the values of x and y obtained in Part B into the original equations and check if both equations are satisfied.
1. For the first equation, x + 6y = 6, substituting x = 3 and y = -1:
3 + 6(-1) = 3 - 6 = -3.
The left side equals -3, which is consistent with the right side.
2. For the second equation, y = 1/3x - 2, substituting x = 3 and y = -1:
-1 = 1/3(3) - 2 = 1 - 2 = -1.
The left side equals -1, which is consistent with the right side.
Both equations are satisfied with x = 3 and y = -1, so algebraically, the solution is verified.
Therefore, the solution to the system of linear equations is x = 3 and y = -1.
We can graph them for you.
The solution will be the place they cross (intersect).