To graph the linear equations \(y_1 = \frac{1}{3}x - 3\) and \(y_2 = -3x + 1\), we will create a table of values for each equation. By selecting different \(x\) values, we can calculate corresponding \(y\) values.
Step 1: Create a Table of Values
For \(y_1 = \frac{1}{3}x - 3\)
| \(x\) | \(y_1 = \frac{1}{3}x - 3\) | |-------|-----------------------------| | -9 | \( \frac{1}{3}(-9) - 3 = -3 - 3 = -6 \) | | 0 | \( \frac{1}{3}(0) - 3 = 0 - 3 = -3 \) | | 3 | \( \frac{1}{3}(3) - 3 = 1 - 3 = -2 \) | | 6 | \( \frac{1}{3}(6) - 3 = 2 - 3 = -1 \) | | 9 | \( \frac{1}{3}(9) - 3 = 3 - 3 = 0 \) |
For \(y_2 = -3x + 1\)
| \(x\) | \(y_2 = -3x + 1\) | |-------|--------------------| | -9 | \( -3(-9) + 1 = 27 + 1 = 28 \) | | 0 | \( -3(0) + 1 = 0 + 1 = 1 \) | | 3 | \( -3(3) + 1 = -9 + 1 = -8 \) | | 6 | \( -3(6) + 1 = -18 + 1 = -17 \) | | 9 | \( -3(9) + 1 = -27 + 1 = -26 \) |
Step 2: Compile the Values
Here are the values we calculated:
For \(y_1\)
- When \(x = -9\), \(y_1 = -6\)
- When \(x = 0\), \(y_1 = -3\)
- When \(x = 3\), \(y_1 = -2\)
- When \(x = 6\), \(y_1 = -1\)
- When \(x = 9\), \(y_1 = 0\)
For \(y_2\)
- When \(x = -9\), \(y_2 = 28\)
- When \(x = 0\), \(y_2 = 1\)
- When \(x = 3\), \(y_2 = -8\)
- When \(x = 6\), \(y_2 = -17\)
- When \(x = 9\), \(y_2 = -26\)
Step 3: Plot the Points
Now, we will plot these points on a coordinate graph:
Points for \(y_1\):
- (-9, -6)
- (0, -3)
- (3, -2)
- (6, -1)
- (9, 0)
Points for \(y_2\):
- (-9, 28)
- (0, 1)
- (3, -8)
- (6, -17)
- (9, -26)
Step 4: Graphing
To graph these, plot each of these points on a Cartesian coordinate system and then connect the points for each line.
Finding the Intersection
The intersection of these two lines is the solution to the system of equations. To find the exact intersection point algebraically, you would set the two equations equal to each other and solve for \(x\) and then find \(y\).
\[ \frac{1}{3}x - 3 = -3x + 1 \]
Multiplying the entire equation by 3 to eliminate the fraction:
\[ x - 9 = -9x + 3 \]
Combine like terms:
\[ 10x = 12 \rightarrow x = \frac{12}{10} = 1.2 \]
Now substitute \(x = 1.2\) back into either equation to find \(y\):
Using \(y_1\):
\[ y_1 = \frac{1}{3}(1.2) - 3 = \frac{1.2}{3} - 3 = 0.4 - 3 = -2.6 \]
So the intersection point (solution) is \((1.2, -2.6)\).
Conclusion
You can now graph these equations and observe where they intersect at the point \((1.2, -2.6)\).