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review the graphs of a system of two linear equations in two variables: x - 5y = 7 and 3x - 2y = -4Asked by 🐠 fish
review the graphs of a system of two linear equations in two variables: x-5y=7 and 3x-2y=-4
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GPT-4o mini
To review the graphs of the system of two linear equations given by:
1. \( x - 5y = 7 \)
2. \( 3x - 2y = -4 \)
We'll start by rewriting each equation in slope-intercept form (y = mx + b) for easier graphing.
### First Equation: \( x - 5y = 7 \)
Rearranging gives:
\[
-5y = -x + 7
\]
Dividing by -5 results in:
\[
y = \frac{1}{5}x - \frac{7}{5}
\]
From this, we can see the slope \( m_1 = \frac{1}{5} \) and the y-intercept \( b_1 = -\frac{7}{5} \).
### Second Equation: \( 3x - 2y = -4 \)
Rearranging gives:
\[
-2y = -3x - 4
\]
Dividing by -2 results in:
\[
y = \frac{3}{2}x + 2
\]
Here, the slope \( m_2 = \frac{3}{2} \) and the y-intercept \( b_2 = 2 \).
### Graphing the Equations
1. **For the first equation \( y = \frac{1}{5}x - \frac{7}{5} \)**:
- The y-intercept is \( -\frac{7}{5} \) (approximately -1.4). This is the point (0, -1.4).
- Using the slope \( \frac{1}{5} \), from (0, -1.4), if we go right 5 units (to x = 5), we go up 1 unit (to y = -1.4 + 1 = -0.4). This gives the point (5, -0.4).
2. **For the second equation \( y = \frac{3}{2}x + 2 \)**:
- The y-intercept is \( 2 \) (the point (0, 2)).
- Using the slope \( \frac{3}{2} \), from (0, 2), if we go right 2 units (to x = 2), we go up 3 units (to y = 2 + 3 = 5). This gives the point (2, 5).
### Finding the Intersection
To find the intersection (the solution to the system), we can solve the equations simultaneously:
From the first equation:
\[
x = 5y + 7
\]
Substituting into the second equation:
\[
3(5y + 7) - 2y = -4
\]
\[
15y + 21 - 2y = -4
\]
\[
13y + 21 = -4
\]
\[
13y = -25
\]
\[
y = -\frac{25}{13}
\]
Substituting back to find \(x\):
\[
x = 5\left(-\frac{25}{13}\right) + 7 = -\frac{125}{13} + \frac{91}{13} = -\frac{34}{13}
\]
### Conclusion
The solution to the system of equations is \( \left(-\frac{34}{13}, -\frac{25}{13}\right) \).
In summary:
- The first equation's graph is a line with a gentle slope that starts below the x-axis.
- The second equation's graph is a steeper line that rises more quickly and crosses the y-axis at 2.
- The two lines intersect at \( \left(-\frac{34}{13}, -\frac{25}{13}\right) \), which represents the solution to the system.
1. \( x - 5y = 7 \)
2. \( 3x - 2y = -4 \)
We'll start by rewriting each equation in slope-intercept form (y = mx + b) for easier graphing.
### First Equation: \( x - 5y = 7 \)
Rearranging gives:
\[
-5y = -x + 7
\]
Dividing by -5 results in:
\[
y = \frac{1}{5}x - \frac{7}{5}
\]
From this, we can see the slope \( m_1 = \frac{1}{5} \) and the y-intercept \( b_1 = -\frac{7}{5} \).
### Second Equation: \( 3x - 2y = -4 \)
Rearranging gives:
\[
-2y = -3x - 4
\]
Dividing by -2 results in:
\[
y = \frac{3}{2}x + 2
\]
Here, the slope \( m_2 = \frac{3}{2} \) and the y-intercept \( b_2 = 2 \).
### Graphing the Equations
1. **For the first equation \( y = \frac{1}{5}x - \frac{7}{5} \)**:
- The y-intercept is \( -\frac{7}{5} \) (approximately -1.4). This is the point (0, -1.4).
- Using the slope \( \frac{1}{5} \), from (0, -1.4), if we go right 5 units (to x = 5), we go up 1 unit (to y = -1.4 + 1 = -0.4). This gives the point (5, -0.4).
2. **For the second equation \( y = \frac{3}{2}x + 2 \)**:
- The y-intercept is \( 2 \) (the point (0, 2)).
- Using the slope \( \frac{3}{2} \), from (0, 2), if we go right 2 units (to x = 2), we go up 3 units (to y = 2 + 3 = 5). This gives the point (2, 5).
### Finding the Intersection
To find the intersection (the solution to the system), we can solve the equations simultaneously:
From the first equation:
\[
x = 5y + 7
\]
Substituting into the second equation:
\[
3(5y + 7) - 2y = -4
\]
\[
15y + 21 - 2y = -4
\]
\[
13y + 21 = -4
\]
\[
13y = -25
\]
\[
y = -\frac{25}{13}
\]
Substituting back to find \(x\):
\[
x = 5\left(-\frac{25}{13}\right) + 7 = -\frac{125}{13} + \frac{91}{13} = -\frac{34}{13}
\]
### Conclusion
The solution to the system of equations is \( \left(-\frac{34}{13}, -\frac{25}{13}\right) \).
In summary:
- The first equation's graph is a line with a gentle slope that starts below the x-axis.
- The second equation's graph is a steeper line that rises more quickly and crosses the y-axis at 2.
- The two lines intersect at \( \left(-\frac{34}{13}, -\frac{25}{13}\right) \), which represents the solution to the system.
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