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

Ask by Frank Mcfarlane. in Canada
Feb 03,2025

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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.

Solución

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.

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Bonus Knowledge

To graph the equation \( y = 3x - 1 \), you can start by plotting the y-intercept at (0, -1). From there, use the slope of 3, which means for every 1 unit you move right along the x-axis, move up 3 units. Draw a straight line through these points, and voilà! For the equation \( 2x - 3y + 1 = 0 \), first rearrange it into slope-intercept form, yielding \( y = \frac{2}{3}x + \frac{1}{3} \). Now, plot the y-intercept at (0, 1/3) and use the slope of \( \frac{2}{3} \) to find additional points. Connect them to create a lovely line!

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