To graph the lines represented by the equations \( y - x = 1 \) and \( 3x - 4y = 12 \), we first need to determine three points that satisfy each equation.
For the first equation: \( y - x = 1 \)
You can rearrange the equation to find \( y \):
\[ y = x + 1 \]
Now, let's pick three values for \( x \) to find corresponding \( y \) values:
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If \( x = 0 \): \[ y = 0 + 1 = 1 \quad \Rightarrow \quad (0, 1) \]
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If \( x = 1 \): \[ y = 1 + 1 = 2 \quad \Rightarrow \quad (1, 2) \]
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If \( x = -1 \): \[ y = -1 + 1 = 0 \quad \Rightarrow \quad (-1, 0) \]
So the three points on the line \( y - x = 1 \) are:
- \( (0, 1) \)
- \( (1, 2) \)
- \( (-1, 0) \)
For the second equation: \( 3x - 4y = 12 \)
First, let's solve for \( y \):
\[ 4y = 3x - 12 \] \[ y = \frac{3}{4}x - 3 \]
Again, we will pick some values for \( x \) to find corresponding \( y \) values:
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If \( x = 0 \): \[ y = \frac{3}{4}(0) - 3 = -3 \quad \Rightarrow \quad (0, -3) \]
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If \( x = 4 \): \[ y = \frac{3}{4}(4) - 3 = 3 - 3 = 0 \quad \Rightarrow \quad (4, 0) \]
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If \( x = 8 \): \[ y = \frac{3}{4}(8) - 3 = 6 - 3 = 3 \quad \Rightarrow \quad (8, 3) \]
So the three points on the line \( 3x - 4y = 12 \) are:
- \( (0, -3) \)
- \( (4, 0) \)
- \( (8, 3) \)
Graphing the Lines
To graph the equations, plot the points you found:
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For \( y - x = 1 \):
- Plot \( (0, 1) \)
- Plot \( (1, 2) \)
- Plot \( (-1, 0) \)
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For \( 3x - 4y = 12 \):
- Plot \( (0, -3) \)
- Plot \( (4, 0) \)
- Plot \( (8, 3) \)
Drawing the Lines
Once the points are plotted, draw a straight line through the points for each equation.
The line for \( y - x = 1 \) will have a positive slope of 1 and will pass through the y-intercept at (0,1).
The line for \( 3x - 4y = 12 \) has a slope of \( \frac{3}{4} \) and will intersect the y-axis at (0, -3).
If you plot these lines on a graphing tool or graph paper, you should see the two lines and where they intersect. If you wish to find the intersection point, you can set the equations equal to one another!