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Sagot :
To classify the given system of equations, we need to determine whether the lines represented by the equations are intersecting, parallel, or coincident. Let's go through the steps to classify the system:
Given the system of equations:
[tex]\[ \begin{aligned} -\frac{1}{2} x & =-2-y \\ 1 + y & = \frac{1}{2} x + 7 \end{aligned} \][/tex]
Step 1: Simplify both equations.
First, we will rewrite each equation in the [tex]\( y = mx + b \)[/tex] form, where [tex]\( m \)[/tex] represents the slope and [tex]\( b \)[/tex] represents the y-intercept.
For the first equation:
[tex]\[ -\frac{1}{2} x = -2 - y \][/tex]
Add [tex]\( y \)[/tex] to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ -\frac{1}{2} x + y = -2 \][/tex]
Add [tex]\(\frac{1}{2} x\)[/tex] to both sides:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
The first equation can be rewritten as:
[tex]\[ y = \frac{1}{2}x - 2 \][/tex]
For the second equation:
[tex]\[ 1 + y = \frac{1}{2} x + 7 \][/tex]
Subtract [tex]\(1 \)[/tex] from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1}{2} x + 6 \][/tex]
The second equation can be rewritten as:
[tex]\[ y = \frac{1}{2}x + 6 \][/tex]
Step 2: Compare the slopes and y-intercepts.
Both equations now have the form [tex]\( y = mx + b \)[/tex], and we can compare the slopes ([tex]\( m \)[/tex]) and y-intercepts ([tex]\( b \)[/tex]):
For both equations:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
[tex]\[ y = \frac{1}{2} x + 6 \][/tex]
We observe the following:
- The slope [tex]\( m \)[/tex] for both equations is [tex]\( \frac{1}{2} \)[/tex].
- The y-intercepts [tex]\( b \)[/tex] are different:
- For the first equation, the y-intercept is [tex]\( -2 \)[/tex].
- For the second equation, the y-intercept is [tex]\( 6 \)[/tex].
Step 3: Determine the relationship.
- Since both equations have the same slope ([tex]\( m = \frac{1}{2} \)[/tex]), the lines are parallel.
- Different y-intercepts indicate that the lines are distinct and do not overlap.
Consequently, the system of equations is classified as parallel.
The correct answer is "parallel".
Given the system of equations:
[tex]\[ \begin{aligned} -\frac{1}{2} x & =-2-y \\ 1 + y & = \frac{1}{2} x + 7 \end{aligned} \][/tex]
Step 1: Simplify both equations.
First, we will rewrite each equation in the [tex]\( y = mx + b \)[/tex] form, where [tex]\( m \)[/tex] represents the slope and [tex]\( b \)[/tex] represents the y-intercept.
For the first equation:
[tex]\[ -\frac{1}{2} x = -2 - y \][/tex]
Add [tex]\( y \)[/tex] to both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ -\frac{1}{2} x + y = -2 \][/tex]
Add [tex]\(\frac{1}{2} x\)[/tex] to both sides:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
The first equation can be rewritten as:
[tex]\[ y = \frac{1}{2}x - 2 \][/tex]
For the second equation:
[tex]\[ 1 + y = \frac{1}{2} x + 7 \][/tex]
Subtract [tex]\(1 \)[/tex] from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1}{2} x + 6 \][/tex]
The second equation can be rewritten as:
[tex]\[ y = \frac{1}{2}x + 6 \][/tex]
Step 2: Compare the slopes and y-intercepts.
Both equations now have the form [tex]\( y = mx + b \)[/tex], and we can compare the slopes ([tex]\( m \)[/tex]) and y-intercepts ([tex]\( b \)[/tex]):
For both equations:
[tex]\[ y = \frac{1}{2} x - 2 \][/tex]
[tex]\[ y = \frac{1}{2} x + 6 \][/tex]
We observe the following:
- The slope [tex]\( m \)[/tex] for both equations is [tex]\( \frac{1}{2} \)[/tex].
- The y-intercepts [tex]\( b \)[/tex] are different:
- For the first equation, the y-intercept is [tex]\( -2 \)[/tex].
- For the second equation, the y-intercept is [tex]\( 6 \)[/tex].
Step 3: Determine the relationship.
- Since both equations have the same slope ([tex]\( m = \frac{1}{2} \)[/tex]), the lines are parallel.
- Different y-intercepts indicate that the lines are distinct and do not overlap.
Consequently, the system of equations is classified as parallel.
The correct answer is "parallel".
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