Answer :
Sure, let's solve the given system of inequalities step-by-step:
Given:
1. [tex]\( x - y > -13 \)[/tex]
2. [tex]\( 2x + y > 0 \)[/tex]
Our objective is to find the region on the coordinate plane that satisfies both inequalities. Here's how we approach solving them:
### Step 1: Convert inequalities to equalities to find boundary lines
1. [tex]\( x - y = -13 \)[/tex]
2. [tex]\( 2x + y = 0 \)[/tex]
### Step 2: Plot the boundary lines
#### Line 1: [tex]\( x - y = -13 \)[/tex]
To graph this line, we find the intercepts:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ 0 - y = -13 \implies y = 13 \][/tex]
- When [tex]\( y = 0 \)[/tex]:
[tex]\[ x - 0 = -13 \implies x = -13 \][/tex]
So, the line passes through points [tex]\((0, 13)\)[/tex] and [tex]\((-13, 0)\)[/tex].
#### Line 2: [tex]\( 2x + y = 0 \)[/tex]
To graph this line, we also find the intercepts:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ 2(0) + y = 0 \implies y = 0 \][/tex]
- When [tex]\( y = 0 \)[/tex]:
[tex]\[ 2x + 0 = 0 \implies x = 0 \][/tex]
In this case, this reduces to [tex]\( y = -2x \)[/tex].
### Step 3: Determine the regions for the inequalities
#### For [tex]\( x - y > -13 \)[/tex]:
Choose a test point not on the line, say [tex]\( (0,0) \)[/tex]:
[tex]\[ 0 - 0 > -13 \implies 0 > -13 \][/tex]
This is true, so the region above the line [tex]\( x - y = -13 \)[/tex] (including the line extending to the right beyond [tex]\(0, 13\)[/tex] and below to the left of [tex]\((-13,0)\)[/tex]) satisfies the inequality.
#### For [tex]\( 2x + y > 0 \)[/tex]:
Again, use the test point [tex]\( (0,0) \)[/tex]:
[tex]\[ 2(0) + 0 > 0 \implies 0 > 0 \][/tex]
This is false; thus, the region above the line [tex]\( y = -2x \)[/tex] or [tex]\(2x + y = 0\)[/tex] (excluding the axis's intersection points and all points below/below-the-line).
### Step 4: Find the intersection region
The solution to the system of inequalities is where both conditions are true. This will be the overlapping region:
1. The region above the line [tex]\( x - y = -13 \)[/tex]
2. The region above the line [tex]\( 2x + y = -0 \)[/tex]or[tex]\( y > -2x \)[/tex]
### Step 5: Graphical representation
Now, plot both lines and shade the regions:
- Line 1: Dashed because it's not included, graph starts at [tex]\( (0,13) \)[/tex] and runs through [tex]\((-13,0)\)[/tex]
- Line 2: Dashed as it's not included, downward-sloped from [tex]\( y = -2x \)[/tex].
Finally, the intersection/visible region where both inequalities overlap represent points verifying both inequalities.
### Conclusion
The required region on the coordinate plane represents points where both inequalities align, described mathematically as:
- [tex]\( x - y > -13 \)[/tex]
- [tex]\( 2x + y > 0 \)[/tex]
Thus, our solution is the set of points that simultaneously satisfy these conditions where both shaded and dashed areas intersect.
Given:
1. [tex]\( x - y > -13 \)[/tex]
2. [tex]\( 2x + y > 0 \)[/tex]
Our objective is to find the region on the coordinate plane that satisfies both inequalities. Here's how we approach solving them:
### Step 1: Convert inequalities to equalities to find boundary lines
1. [tex]\( x - y = -13 \)[/tex]
2. [tex]\( 2x + y = 0 \)[/tex]
### Step 2: Plot the boundary lines
#### Line 1: [tex]\( x - y = -13 \)[/tex]
To graph this line, we find the intercepts:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ 0 - y = -13 \implies y = 13 \][/tex]
- When [tex]\( y = 0 \)[/tex]:
[tex]\[ x - 0 = -13 \implies x = -13 \][/tex]
So, the line passes through points [tex]\((0, 13)\)[/tex] and [tex]\((-13, 0)\)[/tex].
#### Line 2: [tex]\( 2x + y = 0 \)[/tex]
To graph this line, we also find the intercepts:
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ 2(0) + y = 0 \implies y = 0 \][/tex]
- When [tex]\( y = 0 \)[/tex]:
[tex]\[ 2x + 0 = 0 \implies x = 0 \][/tex]
In this case, this reduces to [tex]\( y = -2x \)[/tex].
### Step 3: Determine the regions for the inequalities
#### For [tex]\( x - y > -13 \)[/tex]:
Choose a test point not on the line, say [tex]\( (0,0) \)[/tex]:
[tex]\[ 0 - 0 > -13 \implies 0 > -13 \][/tex]
This is true, so the region above the line [tex]\( x - y = -13 \)[/tex] (including the line extending to the right beyond [tex]\(0, 13\)[/tex] and below to the left of [tex]\((-13,0)\)[/tex]) satisfies the inequality.
#### For [tex]\( 2x + y > 0 \)[/tex]:
Again, use the test point [tex]\( (0,0) \)[/tex]:
[tex]\[ 2(0) + 0 > 0 \implies 0 > 0 \][/tex]
This is false; thus, the region above the line [tex]\( y = -2x \)[/tex] or [tex]\(2x + y = 0\)[/tex] (excluding the axis's intersection points and all points below/below-the-line).
### Step 4: Find the intersection region
The solution to the system of inequalities is where both conditions are true. This will be the overlapping region:
1. The region above the line [tex]\( x - y = -13 \)[/tex]
2. The region above the line [tex]\( 2x + y = -0 \)[/tex]or[tex]\( y > -2x \)[/tex]
### Step 5: Graphical representation
Now, plot both lines and shade the regions:
- Line 1: Dashed because it's not included, graph starts at [tex]\( (0,13) \)[/tex] and runs through [tex]\((-13,0)\)[/tex]
- Line 2: Dashed as it's not included, downward-sloped from [tex]\( y = -2x \)[/tex].
Finally, the intersection/visible region where both inequalities overlap represent points verifying both inequalities.
### Conclusion
The required region on the coordinate plane represents points where both inequalities align, described mathematically as:
- [tex]\( x - y > -13 \)[/tex]
- [tex]\( 2x + y > 0 \)[/tex]
Thus, our solution is the set of points that simultaneously satisfy these conditions where both shaded and dashed areas intersect.
