Solving Linear Inequalities: A Step-by-Step Guide

by Tim Redaksi 50 views
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Hey guys! Today, we're diving into the awesome world of linear inequalities and figuring out how to pinpoint those solution regions that make everything true. It might sound a bit intimidating, but trust me, once you get the hang of it, it's like riding a bike. We'll break it down step by step, so you can confidently tackle any inequality that comes your way. Let's jump right in!

Understanding Linear Inequalities

Before we start plotting and shading, let's make sure we're all on the same page. Linear inequalities are mathematical statements that compare two expressions using inequality symbols like ≤ (less than or equal to), ≥ (greater than or equal to), < (less than), or > (greater than). Unlike equations that have a single solution or a set of discrete solutions, inequalities usually have a range of values that satisfy them. When we're dealing with two variables (like x and y), these solutions form a region on a graph. Our goal is to identify and shade that region.

The key to solving linear inequalities graphically is to first treat the inequality as an equation and graph the corresponding line. This line acts as a boundary, separating the coordinate plane into two regions. One of these regions contains all the points that satisfy the inequality. To figure out which region to shade, we test a point (usually (0,0) if it doesn't lie on the line) in the original inequality. If the point satisfies the inequality, we shade the region containing that point; otherwise, we shade the other region. Remember, if the inequality includes ≤ or ≥, the boundary line is solid, indicating that points on the line are also solutions. If the inequality includes < or >, the boundary line is dashed, indicating that points on the line are not solutions.

Now that we've got the basic concepts down, let's dive into solving some specific inequalities. We'll take each one step by step, so you can see exactly how it's done. By the end of this guide, you'll be a pro at finding those solution regions!

Solving the Inequalities

Let's tackle each inequality one by one, showing you how to find the solution region for each.

a. 3x + 4y ≤ 12

Okay, let's start with the inequality 3x + 4y ≤ 12. Our mission is to find all the points (x, y) that make this statement true. Here's how we do it:

  1. Turn it into an equation: Pretend the ≤ is an = sign. So, we have 3x + 4y = 12. This is the equation of a line.
  2. Graph the line: To graph the line, find two points on it. A simple way to do this is to find the x and y intercepts. Set x = 0 and solve for y: 3(0) + 4y = 12 => y = 3. So, the point (0, 3) is on the line. Now set y = 0 and solve for x: 3x + 4(0) = 12 => x = 4. So, the point (4, 0) is on the line. Plot these two points and draw a line through them. Since we have ≤ (less than or equal to), the line will be solid.
  3. Pick a test point: Choose a point that's not on the line. The easiest point is usually (0, 0). Plug it into the original inequality: 3(0) + 4(0) ≤ 12 => 0 ≤ 12. This is true!
  4. Shade the correct region: Since (0, 0) made the inequality true, we shade the region that includes (0, 0). This is the region below the line.

So, the solution region for 3x + 4y ≤ 12 is the area on and below the line 3x + 4y = 12. Any point in that shaded region will satisfy the inequality.

b. 2x + 3y ≥ -6

Next up, we have 2x + 3y ≥ -6. Let's follow the same steps as before:

  1. Turn it into an equation: 2x + 3y = -6
  2. Graph the line: Find the intercepts. Set x = 0: 2(0) + 3y = -6 => y = -2. So, (0, -2) is on the line. Set y = 0: 2x + 3(0) = -6 => x = -3. So, (-3, 0) is on the line. Plot these points and draw a solid line through them (because of the ≥).
  3. Pick a test point: Again, let's use (0, 0). Plug it into the original inequality: 2(0) + 3(0) ≥ -6 => 0 ≥ -6. This is true!
  4. Shade the correct region: Since (0, 0) works, we shade the region containing (0, 0), which is the area above the line.

The solution region for 2x + 3y ≥ -6 is the area on and above the line 2x + 3y = -6.

c. x - 4y ≤ 4

Moving on to x - 4y ≤ 4:

  1. Turn it into an equation: x - 4y = 4
  2. Graph the line: Find the intercepts. Set x = 0: 0 - 4y = 4 => y = -1. So, (0, -1) is on the line. Set y = 0: x - 4(0) = 4 => x = 4. So, (4, 0) is on the line. Plot these points and draw a solid line.
  3. Pick a test point: Use (0, 0): 0 - 4(0) ≤ 4 => 0 ≤ 4. This is true!
  4. Shade the correct region: Since (0, 0) works, we shade the region containing (0, 0), which is the area above the line.

The solution region for x - 4y ≤ 4 is the area on and above the line x - 4y = 4.

d. 2x - 8 ≤ 0

Now let's solve 2x - 8 ≤ 0. This one is a bit simpler because it only involves x:

  1. Solve for x: 2x - 8 ≤ 0 => 2x ≤ 8 => x ≤ 4
  2. Graph the line: This is a vertical line at x = 4. Draw a solid vertical line at x = 4.
  3. Pick a test point: Choose a point to the left or right of the line. Let's use (0, 0): 2(0) - 8 ≤ 0 => -8 ≤ 0. This is true!
  4. Shade the correct region: Since (0, 0) works, we shade the region to the left of the line.

The solution region for 2x - 8 ≤ 0 is the area on and to the left of the vertical line x = 4.

e. 4y + 12 ≥ 0

Next, we have 4y + 12 ≥ 0. This one only involves y:

  1. Solve for y: 4y + 12 ≥ 0 => 4y ≥ -12 => y ≥ -3
  2. Graph the line: This is a horizontal line at y = -3. Draw a solid horizontal line at y = -3.
  3. Pick a test point: Choose a point above or below the line. Let's use (0, 0): 4(0) + 12 ≥ 0 => 12 ≥ 0. This is true!
  4. Shade the correct region: Since (0, 0) works, we shade the region above the line.

The solution region for 4y + 12 ≥ 0 is the area on and above the horizontal line y = -3.

f. x + y ≥ 0

Let's tackle x + y ≥ 0:

  1. Turn it into an equation: x + y = 0
  2. Graph the line: This line passes through the origin (0, 0). To find another point, let x = 1. Then 1 + y = 0 => y = -1. So, (1, -1) is on the line. Plot these points and draw a solid line.
  3. Pick a test point: We can't use (0, 0) because it's on the line. Let's use (1, 0): 1 + 0 ≥ 0 => 1 ≥ 0. This is true!
  4. Shade the correct region: Since (1, 0) works, we shade the region containing (1, 0), which is the area above the line.

The solution region for x + y ≥ 0 is the area on and above the line x + y = 0.

g. x - y ≤ 0

Finally, we have x - y ≤ 0:

  1. Turn it into an equation: x - y = 0
  2. Graph the line: This line also passes through the origin (0, 0). Let x = 1. Then 1 - y = 0 => y = 1. So, (1, 1) is on the line. Plot these points and draw a solid line.
  3. Pick a test point: Again, we can't use (0, 0). Let's use (1, 0): 1 - 0 ≤ 0 => 1 ≤ 0. This is false!
  4. Shade the correct region: Since (1, 0) doesn't work, we shade the other region, which is the area above the line.

The solution region for x - y ≤ 0 is the area on and above the line x - y = 0.

Conclusion

And there you have it! We've walked through each inequality step by step, showing you how to find and shade the solution regions. Remember, the key is to turn the inequality into an equation, graph the line, pick a test point, and shade the correct region. With a little practice, you'll be solving linear inequalities like a pro. Keep up the great work, and happy graphing!