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Graph the linear equations using a table of values.

y(1) =1/3x - 3
y(2) = -3x + 1

x
y(1)=
1/3x-3
y(2)=
1/3x + 1
(x,y(1))
(x, y(2))
3

6

-9

Graph the equations to see the solution.

3 answers

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 \((1.2, -2.6)\).

Could you put it in the table

Sure! Here are the values for both equations laid out in a clear table format.

Table of Values for \(y_1 = \frac{1}{3}x - 3\) and \(y_2 = -3x + 1\)

| \(x\) | \(y_1 = \frac{1}{3}x - 3\) | \(y_2 = -3x + 1\) | |-------|-----------------------------|--------------------| | -9 | -6 | 28 | | 0 | -3 | 1 | | 3 | -2 | -8 | | 6 | -1 | -17 | | 9 | 0 | -26 | | 1.2 | -2.6 | -2.6 |

Notes:

  • The last row (\(x = 1.2\)) shows the intersection point of the two equations. At \(x = 1.2\), both \(y_1\) and \(y_2\) yield \(y = -2.6\).
  • This demonstrates the solution to the system of equations graphically: the point where both lines intersect.

You can use this table to plot the points on a graph for visualization.