Solved RISCvia rutes 2.8 - Graph Linear Inequalities in Two ... - Free Printable

Problem Analysis and Solution

The worksheet focuses on graphing linear inequalities in two variables. Let's solve each part step by step.

---

#### 1) Which ordered pair is a solution of \( 5x - 2y \leq 6 \)?

We need to substitute each ordered pair into the inequality \( 5x - 2y \leq 6 \) and check which one satisfies it.

- Option A: (3, 3)
\[
5(3) - 2(3) = 15 - 6 = 9
\]
Since \( 9 \not\leq 6 \), (3, 3) is not a solution.

- Option B: (5, 5)
\[
5(5) - 2(5) = 25 - 10 = 15
\]
Since \( 15 \not\leq 6 \), (5, 5) is not a solution.

- Option C: (1, -2)
\[
5(1) - 2(-2) = 5 + 4 = 9
\]
Since \( 9 \not\leq 6 \), (1, -2) is not a solution.

- Option D: (0, -3)
\[
5(0) - 2(-3) = 0 + 6 = 6
\]
Since \( 6 \leq 6 \), (0, -3) is a solution.

Answer to Question 1:
\[
\boxed{D}
\]

---

#### 2) Graph the following inequalities in a coordinate plane.

We will graph each inequality step by step.

##### a) \( x \leq 4 \)

- This inequality represents all points where the \( x \)-coordinate is less than or equal to 4.
- Draw a vertical line at \( x = 4 \). Since the inequality includes "≤", the line is solid.
- Shade the region to the left of the line (where \( x \leq 4 \)).

##### b) \( y > -2 \)

- This inequality represents all points where the \( y \)-coordinate is greater than -2.
- Draw a horizontal line at \( y = -2 \). Since the inequality does not include "≤", the line is dashed.
- Shade the region above the line (where \( y > -2 \)).

##### c) \( 3x - 4y > 12 \)

1. Rewrite the inequality as an equation to find the boundary line:
\[
3x - 4y = 12
\]
2. Find the intercepts:
- When \( x = 0 \):
\[
3(0) - 4y = 12 \implies -4y = 12 \implies y = -3
\]
Point: \( (0, -3) \)
- When \( y = 0 \):
\[
3x - 4(0) = 12 \implies 3x = 12 \implies x = 4
\]
Point: \( (4, 0) \)
3. Plot the points \( (0, -3) \) and \( (4, 0) \) and draw a dashed line through them (since the inequality is strict, \( > \)).
4. Test a point not on the line (e.g., \( (0, 0) \)):
\[
3(0) - 4(0) > 12 \implies 0 > 12 \quad \text{(False)}
\]
So, shade the region above the line.

##### d) \( y \geq \frac{3}{2}x + 3 \)

1. Rewrite the inequality as an equation to find the boundary line:
\[
y = \frac{3}{2}x + 3
\]
2. Find the intercepts:
- When \( x = 0 \):
\[
y = \frac{3}{2}(0) + 3 = 3
\]
Point: \( (0, 3) \)
- When \( y = 0 \):
\[
0 = \frac{3}{2}x + 3 \implies \frac{3}{2}x = -3 \implies x = -2
\]
Point: \( (-2, 0) \)
3. Plot the points \( (0, 3) \) and \( (-2, 0) \) and draw a solid line through them (since the inequality includes "≥").
4. Test a point not on the line (e.g., \( (0, 0) \)):
\[
0 \geq \frac{3}{2}(0) + 3 \implies 0 \geq 3 \quad \text{(False)}
\]
So, shade the region above the line.

##### e) \( y > -|x - 1| + 2 \)

1. Analyze the absolute value function:
\[
y = -|x - 1| + 2
\]
- The vertex of the V-shaped graph occurs at \( x = 1 \).
- At \( x = 1 \):
\[
y = -|1 - 1| + 2 = 2
\]
Vertex: \( (1, 2) \)
- For \( x < 1 \):
\[
y = -(1 - x) + 2 = x + 1
\]
- For \( x > 1 \):
\[
y = -(x - 1) + 2 = -x + 3
\]
2. Graph the V-shaped boundary:
- Plot the lines \( y = x + 1 \) (for \( x < 1 \)) and \( y = -x + 3 \) (for \( x > 1 \)).
- The vertex is at \( (1, 2) \).
3. Since the inequality is strict (\( > \)), the boundary lines are dashed.
4. Test a point not on the boundary (e.g., \( (0, 0) \)):
\[
0 > -|0 - 1| + 2 \implies 0 > -1 + 2 \implies 0 > 1 \quad \text{(False)}
\]
So, shade the region above the V-shaped boundary.

##### f) \( y \geq 2|x + 2| - 1 \)

1. Analyze the absolute value function:
\[
y = 2|x + 2| - 1
\]
- The vertex of the V-shaped graph occurs at \( x = -2 \).
- At \( x = -2 \):
\[
y = 2|-2 + 2| - 1 = -1
\]
Vertex: \( (-2, -1) \)
- For \( x < -2 \):
\[
y = 2(-(x + 2)) - 1 = -2x - 4 - 1 = -2x - 5
\]
- For \( x > -2 \):
\[
y = 2(x + 2) - 1 = 2x + 4 - 1 = 2x + 3
\]
2. Graph the V-shaped boundary:
- Plot the lines \( y = -2x - 5 \) (for \( x < -2 \)) and \( y = 2x + 3 \) (for \( x > -2 \)).
- The vertex is at \( (-2, -1) \).
3. Since the inequality includes "≥", the boundary lines are solid.
4. Test a point not on the boundary (e.g., \( (0, 0) \)):
\[
0 \geq 2|0 + 2| - 1 \implies 0 \geq 2(2) - 1 \implies 0 \geq 4 - 1 \implies 0 \geq 3 \quad \text{(False)}
\]
So, shade the region above the V-shaped boundary.

---

Final Answer:

\[
\boxed{D}
\]