Answer :
Certainly! Let's graph the system of inequalities:
[tex]\[ \begin{array}{l} y \leq 2x + 1 \\ y < -x - 1 \end{array} \][/tex]
To graph these inequalities, we can follow these steps:
1. Graph the line [tex]\( y = 2x + 1 \)[/tex]:
- This is the boundary line for the inequality [tex]\( y \leq 2x + 1 \)[/tex].
- To graph this line, we need at least two points.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0) + 1 = 1 \quad \text{(Point: (0, 1))} \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1) + 1 = 3 \quad \text{(Point: (1, 3))} \][/tex]
- Plot these points and draw a straight line through them. Since the inequality is [tex]\( y \leq 2x + 1 \)[/tex], we will shade the region below this line (including the line itself because of the [tex]\(\leq\)[/tex]).
2. Graph the line [tex]\( y = -x - 1 \)[/tex]:
- This is the boundary line for the inequality [tex]\( y < -x - 1 \)[/tex].
- To graph this line, we need at least two points.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -0 - 1 = -1 \quad \text{(Point: (0, -1))} \][/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -(-1) - 1 = 0 \quad \text{(Point: (-1, 0))} \][/tex]
- Plot these points and draw a dashed line through them. Since the inequality is [tex]\( y < -x - 1 \)[/tex], we will shade the region below this line without including the line itself (indicated by the '<').
3. Find the intersection of the two lines:
- To find the intersection point, set [tex]\( y = 2x + 1 \)[/tex] equal to [tex]\( y = -x - 1 \)[/tex]:
[tex]\[ 2x + 1 = -x - 1 \][/tex]
[tex]\[ 3x = -2 \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
- Plug this value of [tex]\( x \)[/tex] back into either equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 2\left(-\frac{2}{3}\right) + 1 = -\frac{4}{3} + 1 = -\frac{1}{3} \][/tex]
- So, the intersection point is [tex]\(\left(-\frac{2}{3}, -\frac{1}{3}\right)\)[/tex].
4. Shading the appropriate regions:
- Shade the region below the line [tex]\( y = 2x + 1 \)[/tex] that includes the line itself.
- Shade the region below the line [tex]\( y = -x - 1 \)[/tex] that does not include the line.
- The solution to the system of inequalities is the region where the shaded areas overlap.
By plotting these on the same graph, we get:
1. Start with the axes, then plot the two points [tex]\((0, 1)\)[/tex] and [tex]\((1, 3)\)[/tex] for the line [tex]\( y = 2x + 1 \)[/tex]. Draw the solid line through these points and shade below it.
2. Next, plot the two points [tex]\((0, -1)\)[/tex] and [tex]\((-1, 0)\)[/tex] for the line [tex]\( y = -x - 1 \)[/tex]. Draw the dashed line through these points and shade below it.
The solution region is where these two shaded areas overlap, which should be a wedge-shaped region that lies below both lines but does not include the dashed line [tex]\( y = -x - 1 \)[/tex].
This intersection region visually represents all the points [tex]\((x, y)\)[/tex] that satisfy both inequalities simultaneously.
[tex]\[ \begin{array}{l} y \leq 2x + 1 \\ y < -x - 1 \end{array} \][/tex]
To graph these inequalities, we can follow these steps:
1. Graph the line [tex]\( y = 2x + 1 \)[/tex]:
- This is the boundary line for the inequality [tex]\( y \leq 2x + 1 \)[/tex].
- To graph this line, we need at least two points.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0) + 1 = 1 \quad \text{(Point: (0, 1))} \][/tex]
- When [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2(1) + 1 = 3 \quad \text{(Point: (1, 3))} \][/tex]
- Plot these points and draw a straight line through them. Since the inequality is [tex]\( y \leq 2x + 1 \)[/tex], we will shade the region below this line (including the line itself because of the [tex]\(\leq\)[/tex]).
2. Graph the line [tex]\( y = -x - 1 \)[/tex]:
- This is the boundary line for the inequality [tex]\( y < -x - 1 \)[/tex].
- To graph this line, we need at least two points.
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -0 - 1 = -1 \quad \text{(Point: (0, -1))} \][/tex]
- When [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -(-1) - 1 = 0 \quad \text{(Point: (-1, 0))} \][/tex]
- Plot these points and draw a dashed line through them. Since the inequality is [tex]\( y < -x - 1 \)[/tex], we will shade the region below this line without including the line itself (indicated by the '<').
3. Find the intersection of the two lines:
- To find the intersection point, set [tex]\( y = 2x + 1 \)[/tex] equal to [tex]\( y = -x - 1 \)[/tex]:
[tex]\[ 2x + 1 = -x - 1 \][/tex]
[tex]\[ 3x = -2 \][/tex]
[tex]\[ x = -\frac{2}{3} \][/tex]
- Plug this value of [tex]\( x \)[/tex] back into either equation to find [tex]\( y \)[/tex]:
[tex]\[ y = 2\left(-\frac{2}{3}\right) + 1 = -\frac{4}{3} + 1 = -\frac{1}{3} \][/tex]
- So, the intersection point is [tex]\(\left(-\frac{2}{3}, -\frac{1}{3}\right)\)[/tex].
4. Shading the appropriate regions:
- Shade the region below the line [tex]\( y = 2x + 1 \)[/tex] that includes the line itself.
- Shade the region below the line [tex]\( y = -x - 1 \)[/tex] that does not include the line.
- The solution to the system of inequalities is the region where the shaded areas overlap.
By plotting these on the same graph, we get:
1. Start with the axes, then plot the two points [tex]\((0, 1)\)[/tex] and [tex]\((1, 3)\)[/tex] for the line [tex]\( y = 2x + 1 \)[/tex]. Draw the solid line through these points and shade below it.
2. Next, plot the two points [tex]\((0, -1)\)[/tex] and [tex]\((-1, 0)\)[/tex] for the line [tex]\( y = -x - 1 \)[/tex]. Draw the dashed line through these points and shade below it.
The solution region is where these two shaded areas overlap, which should be a wedge-shaped region that lies below both lines but does not include the dashed line [tex]\( y = -x - 1 \)[/tex].
This intersection region visually represents all the points [tex]\((x, y)\)[/tex] that satisfy both inequalities simultaneously.
