Graphs and Lines 1. Graph each lis (a) $$ y=3 x-1 $$ c) $$ 2 x-3 y+1 $$

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It seems that I cannot plot the graphs directly at the moment. However, I can guide you through the process of graphing the equations step by step.

### 1. Graphing $$ y = 3x - 1 $$

This is a linear equation in slope-intercept form $$ y = mx + b $$, where:
- $$ m = 3 $$ (the slope)
- $$ b = -1 $$ (the y-intercept)

**Steps to graph:**
- **Find the y-intercept:** Set $$ x = 0 $$:
  $$
  y = 3(0) - 1 = -1
  $$
  So, the point is $$ (0, -1) $$.

- **Find another point using the slope:** The slope $$ m = 3 $$ means that for every 1 unit you move to the right (increase $$ x $$ by 1), you move up 3 units (increase $$ y $$ by 3).
  - Starting from $$ (0, -1) $$, if $$ x = 1 $$:
  $$
  y = 3(1) - 1 = 2
  $$
  So, another point is $$ (1, 2) $$.

- **Plot the points:** Plot $$ (0, -1) $$ and $$ (1, 2) $$ on the graph and draw a line through them.

### 2. Graphing $$ 2x - 3y + 1 = 0 $$

First, we can rearrange this equation into slope-intercept form $$ y = mx + b $$.

**Rearranging the equation:**
$$
2x - 3y + 1 = 0 \implies -3y = -2x - 1 \implies y = \frac{2}{3}x + \frac{1}{3}
$$

Now, we have:
- Slope $$ m = \frac{2}{3} $$
- Y-intercept $$ b = \frac{1}{3} $$

**Steps to graph:**
- **Find the y-intercept:** Set $$ x = 0 $$:
  $$
  y = \frac{2}{3}(0) + \frac{1}{3} = \frac{1}{3}
  $$
  So, the point is $$ (0, \frac{1}{3}) $$.

- **Find another point using the slope:** The slope $$ m = \frac{2}{3} $$ means that for every 3 units you move to the right (increase $$ x $$ by 3), you move up 2 units (increase $$ y $$ by 2).
  - Starting from $$ (0, \frac{1}{3}) $$, if $$ x = 3 $$:
  $$
  y = \frac{2}{3}(3) + \frac{1}{3} = 2 + \frac{1}{3} = \frac{7}{3}
  $$
  So, another point is $$ (3, \frac{7}{3}) $$.

- **Plot the points:** Plot $$ (0, \frac{1}{3}) $$ and $$ (3, \frac{7}{3}) $$ on the graph and draw a line through them.

### Summary
- For $$ y = 3x - 1 $$, plot points $$ (0, -1) $$ and $$ (1, 2) $$.
- For $$ 2x - 3y + 1 = 0 $$, plot points $$ (0, \frac{1}{3}) $$ and $$ (3, \frac{7}{3}) $$.

You can use graph paper or a graphing tool to visualize these lines based on the points calculated.
To graph the equations: 1. **For $$ y = 3x - 1 $$:** - Plot the y-intercept at $$ (0, -1) $$. - Use the slope of 3 to find another point, such as $$ (1, 2) $$. - Draw a straight line through these points. 2. **For $$ 2x - 3y + 1 = 0 $$:** - Rearrange to $$ y = \frac{2}{3}x + \frac{1}{3} $$. - Plot the y-intercept at $$ (0, \frac{1}{3}) $$. - Use the slope of $$ \frac{2}{3} $$ to find another point, such as $$ (3, \frac{7}{3}) $$. - Draw a straight line through these points. This will give you the graphs of both equations.

Graphs and Lines 1. Graph each lis (a) \( y=3 x-1 \) c) \( 2 x-3 y+1 \)

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