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

Let's solve the problems step by step.

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Problem 1:

Which ordered pair is a solution of $ 5x - 2y \leq 6 $?

We are given four choices:

A) (3, 3)
B) (5, 5)
C) (1, -2)
D) (0, -3)

We will substitute each ordered pair into the inequality $ 5x - 2y \leq 6 $ and check whether it satisfies the inequality.

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#### A) (3, 3)
$ 5(3) - 2(3) = 15 - 6 = 9 $
Is $ 9 \leq 6 $? → No

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#### B) (5, 5)
$ 5(5) - 2(5) = 25 - 10 = 15 $
Is $ 15 \leq 6 $? → No

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#### C) (1, -2)
$ 5(1) - 2(-2) = 5 + 4 = 9 $
Is $ 9 \leq 6 $? → No

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#### D) (0, -3)
$ 5(0) - 2(-3) = 0 + 6 = 6 $
Is $ 6 \leq 6 $? → Yes!

✔ So, D) (0, -3) is a solution.

> ✔ Answer: D) (0, -3)

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Problem 2: Graph the following inequalities in a coordinate plane.

We'll go through each one and describe how to graph them. Since I can't draw on the image, I’ll explain the steps clearly so you can graph them accurately.

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#### a) $ x \leq 4 $

- This is a vertical line at $ x = 4 $.
- Since it's $ \leq $, we shade to the left of the line (including the line).
- Draw a solid vertical line at $ x = 4 $, then shade all points where $ x \leq 4 $.

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#### b) $ y > -2 $

- This is a horizontal line at $ y = -2 $.
- Since it's $ > $, we shade above the line (not including the line).
- Draw a dashed horizontal line at $ y = -2 $, then shade the region above it.

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#### c) $ 3x - 4y > 12 $

First, rewrite in slope-intercept form:
$$
3x - 4y > 12 \\
-4y > -3x + 12 \\
y < \frac{3}{4}x - 3 \quad \text{(Divide by -4; reverse inequality)}
$$

- Graph the line $ y = \frac{3}{4}x - 3 $
- Use a dashed line because it's strict inequality ($ > $)
- Shade below the line (since $ y < \ldots $)

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

- This is already in slope-intercept form.
- Graph the line $ y = \frac{2}{3}x + 3 $
- Use a solid line because it's $ \geq $
- Shade above the line (since $ y \geq \ldots $)

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#### e) $ y > -|x - 1| + 2 $

This is an absolute value inequality.

1. First, graph the parent function: $ y = -|x - 1| + 2 $

- The vertex is at $ (1, 2) $
- The negative sign makes it open downward
- It's a V-shape with peak at (1, 2)

2. Since the inequality is $ y > $, we:
- Draw the graph as a dashed V-shape (because it's strict inequality)
- Shade the region above the V-shape

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

Again, absolute value.

1. Graph $ y = 2|x + 2| - 1 $

- Vertex: when $ x + 2 = 0 $ → $ x = -2 $, so vertex is at $ (-2, -1) $
- The coefficient 2 makes it narrower than the basic |x|
- Opens upward

2. Inequality: $ y \geq $ → use a solid V-shape
- Shade above the V-shape

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✔ Summary of Graphing Steps:

| Inequality | Line Type | Shade |
|----------|-----------|-------|
| a) $ x \leq 4 $ | Solid vertical line at $ x=4 $ | Left |
| b) $ y > -2 $ | Dashed horizontal line at $ y=-2 $ | Above |
| c) $ 3x - 4y > 12 $ | Dashed line $ y = \frac{3}{4}x - 3 $ | Below |
| d) $ y \geq \frac{2}{3}x + 3 $ | Solid line $ y = \frac{2}{3}x + 3 $ | Above |
| e) $ y > -|x - 1| + 2 $ | Dashed V-shape, vertex (1,2), opens down | Above |
| f) $ y \geq 2|x + 2| - 1 $ | Solid V-shape, vertex (-2,-1), opens up | Above |

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Let me know if you'd like me to generate a visual description or sketch for any of these!