Mastering Inequalities: Solving For Rational Numbers

Welcome, math enthusiasts! Today, we're diving deep into the fascinating world of inequalities, specifically focusing on how to solve them within the domain of rational numbers (Q). Inequalities are fundamental tools in mathematics, allowing us to express relationships where quantities are not necessarily equal but fall within a certain range. Understanding how to manipulate and solve these is crucial for various mathematical and real-world applications. We'll be tackling three distinct problems, each designed to test and enhance your understanding of inequality properties and rational number operations. So, grab your notebooks, sharpen your pencils, and let's embark on this mathematical journey together!

Understanding Rational Numbers and Inequalities

Before we jump into solving, let's quickly refresh what we mean by rational numbers (Q). Simply put, rational numbers are any numbers that can be expressed as a fraction p/q, where p and q are integers, and q is not zero. This includes all integers, terminating decimals, and repeating decimals. The domain of rational numbers is rich and expansive, and it's within this domain that we'll be finding the solutions to our inequalities. Now, what about inequalities? Unlike equations that state equality (e.g., x = 5), inequalities express relationships of 'less than' (<), 'greater than' (>), 'less than or equal to' (≤), or 'greater than or equal to' (≥). The key to solving inequalities lies in maintaining the balance of the inequality sign while isolating the variable. Most of the rules for solving inequalities are similar to those for solving equations, with one crucial exception: multiplying or dividing both sides by a negative number reverses the inequality sign. This is a fundamental rule that we must always keep in mind. We'll be applying these principles systematically to each problem, ensuring we arrive at the correct solution set for y or x within the realm of rational numbers. Remember, the goal is to isolate the variable on one side of the inequality, much like solving an equation, but with careful attention to the direction of the inequality sign. Our journey will involve basic arithmetic, fraction manipulation, and a solid grasp of algebraic principles. Let's get started with our first problem, which involves fractions and variables on both sides, a common scenario in inequality problems.

Problem 3.5.a: Solving 2y - 3 < ½ (7 - y)

Let's begin our exploration of solving inequalities with the first problem: 2y - 3 < ½ (7 - y). Our primary objective here is to isolate the variable y on one side of the inequality sign. The presence of a fraction and parentheses suggests our first steps should be to simplify the expression. We'll start by distributing the ½ to the terms inside the parentheses on the right side. This gives us: 2y - 3 < (½ * 7) - (½ * y), which simplifies to 2y - 3 < 3.5 - ½y. Now that the parentheses are gone, we want to gather all terms containing y on one side and all constant terms on the other. Let's add ½y to both sides of the inequality. Remember, adding or subtracting terms does not change the direction of the inequality sign. So, we have: 2y + ½y - 3 < 3.5. Combining the y terms: (2 + ½)y - 3 < 3.5, which is (4/2 + 1/2)y - 3 < 3.5, or (5/2)y - 3 < 3.5. Next, we'll move the constant term (-3) to the right side by adding 3 to both sides: (5/2)y < 3.5 + 3, resulting in (5/2)y < 6.5. To finally isolate y, we need to divide both sides by 5/2. Dividing by a positive number does not change the inequality sign. Remember that dividing by a fraction is the same as multiplying by its reciprocal. So, we multiply both sides by 2/5: y < 6.5 * (2/5). Let's convert 6.5 to a fraction to make multiplication easier: 6.5 = 13/2. So, the inequality becomes: y < (13/2) * (2/5). The 2s cancel out, leaving us with y < 13/5. In decimal form, this is y < 2.6. Therefore, the solution to the inequality 2y - 3 < ½ (7 - y) in the domain of rational numbers is all rational numbers y such that y < 13/5 or y < 2.6. This means any rational number less than 2.6 will satisfy the original inequality. It's a good practice to test a value, say y = 0, which is less than 2.6. Substituting into the original inequality: 2(0) - 3 < ½ (7 - 0) => -3 < ½(7) => -3 < 3.5, which is true. Let's test a value not in the solution set, say y = 3, which is greater than 2.6. Substituting: 2(3) - 3 < ½ (7 - 3) => 6 - 3 < ½(4) => 3 < 2, which is false. This confirms our solution.

Problem 3.5.b: Solving (x - 2) + 4(2x + 1) ≥ 4(x - 3)

Now, let's tackle our second inequality: (x - 2) + 4(2x + 1) ≥ 4(x - 3). This problem involves multiple sets of parentheses and terms that need simplification before we can isolate the variable x. We'll start by distributing the constants outside the parentheses to the terms within. On the left side, we distribute the 4 to (2x + 1): (x - 2) + (4 * 2x) + (4 * 1) ≥ 4(x - 3), which becomes x - 2 + 8x + 4 ≥ 4(x - 3). Now, let's combine like terms on the left side: (x + 8x) + (-2 + 4) ≥ 4(x - 3), giving us 9x + 2 ≥ 4(x - 3). On the right side, we distribute the 4 to (x - 3): 9x + 2 ≥ (4 * x) - (4 * 3), resulting in 9x + 2 ≥ 4x - 12. Our next step is to gather all x terms on one side and constant terms on the other. Let's subtract 4x from both sides. Remember, subtracting a term does not alter the inequality sign: 9x - 4x + 2 ≥ 4x - 4x - 12, which simplifies to 5x + 2 ≥ -12. Now, we'll move the constant term (+2) to the right side by subtracting 2 from both sides: 5x + 2 - 2 ≥ -12 - 2, leading to 5x ≥ -14. Finally, to isolate x, we divide both sides by 5. Since 5 is a positive number, the inequality sign remains the same: x ≥ -14/5. In decimal form, this is x ≥ -2.8. Thus, the solution to the inequality (x - 2) + 4(2x + 1) ≥ 4(x - 3) in the domain of rational numbers is all rational numbers x such that x ≥ -14/5 or x ≥ -2.8. This means any rational number greater than or equal to -2.8 will satisfy the original inequality. Let's test a value. We'll pick x = 0, which is greater than -2.8. Substituting into the original inequality: (0 - 2) + 4(20 + 1) ≥ 4(0 - 3) => -2 + 4(1) ≥ 4(-3) => -2 + 4 ≥ -12 => 2 ≥ -12, which is true. Now, let's test a value not in the solution set, say x = -3, which is less than -2.8. Substituting: (-3 - 2) + 4(2(-3) + 1) ≥ 4(-3 - 3) => -5 + 4(-6 + 1) ≥ 4(-6) => -5 + 4(-5) ≥ -24 => -5 - 20 ≥ -24 => -25 ≥ -24, which is false. This verification confirms our solution set.

Problem 3.5.c: Solving ½ (3x - ⅔) + 6 ≤ x + 5

Our final inequality for today is ½ (3x - ⅔) + 6 ≤ x + 5. This problem also features fractions and requires careful algebraic manipulation. We begin by distributing the ½ to the terms within the parentheses on the left side: (½ * 3x) - (½ * ⅔) + 6 ≤ x + 5. This simplifies to (3/2)x - 1/3 + 6 ≤ x + 5. Let's combine the constant terms on the left side: -1/3 + 6. To do this, we find a common denominator, which is 3. So, -1/3 + 18/3 = 17/3. The inequality now reads: (3/2)x + 17/3 ≤ x + 5. Our goal is to gather all x terms on one side and constants on the other. Let's subtract x from both sides. Since x is the same as 2/2 x, we have: (3/2)x - (2/2)x + 17/3 ≤ x - x + 5, which simplifies to (1/2)x + 17/3 ≤ 5. Now, we move the constant term (17/3) to the right side by subtracting 17/3 from both sides: (1/2)x ≤ 5 - 17/3. To perform the subtraction on the right side, we find a common denominator, which is 3. So, 5 = 15/3. Thus, (1/2)x ≤ 15/3 - 17/3, resulting in (1/2)x ≤ -2/3. To isolate x, we need to multiply both sides by 2 (the reciprocal of 1/2). Multiplying by a positive number does not change the inequality sign: x ≤ (-2/3) * 2. This gives us x ≤ -4/3. Therefore, the solution to the inequality ½ (3x - ⅔) + 6 ≤ x + 5 in the domain of rational numbers is all rational numbers x such that x ≤ -4/3. This means any rational number less than or equal to -4/3 will satisfy the original inequality. Let's check our work with a value. We'll choose x = -2, which is less than -4/3 (since -2 = -6/3). Substituting into the original inequality: ½ (3(-2) - ⅔) + 6 ≤ -2 + 5 => ½ (-6 - ⅔) + 6 ≤ 3. Let's simplify the term in the parentheses: -6 = -18/3. So, -18/3 - 2/3 = -20/3. Now we have: ½ (-20/3) + 6 ≤ 3 => -10/3 + 6 ≤ 3. Convert 6 to thirds: 18/3. So, -10/3 + 18/3 ≤ 3 => 8/3 ≤ 3. Since 3 = 9/3, the inequality becomes 8/3 ≤ 9/3, which is true. Let's test a value not in the solution set, say x = 0, which is greater than -4/3. Substituting: ½ (3(0) - ⅔) + 6 ≤ 0 + 5 => ½ (-⅔) + 6 ≤ 5 => -1/3 + 6 ≤ 5. Convert 6 to thirds: 18/3. So, -1/3 + 18/3 ≤ 5 => 17/3 ≤ 5. Since 5 = 15/3, the inequality is 17/3 ≤ 15/3, which is false. Our solution is correct!

Conclusion: The Power of Solving Inequalities

We've successfully navigated through three distinct inequalities, solving each within the domain of rational numbers (Q). You've seen how to handle fractions, distribute terms, combine like terms, and most importantly, how to isolate a variable while respecting the rules of inequalities. Remember the golden rule: multiplying or dividing by a negative number reverses the inequality sign. Each step we took was deliberate, aiming to simplify the expression and isolate the variable without altering the fundamental truth of the inequality. Whether it was finding y < 2.6, x ≥ -2.8, or x ≤ -4/3, the process involved a systematic application of algebraic principles. These skills are not just for textbook problems; they are foundational for understanding functions, graphing, optimization problems in calculus, and countless real-world scenarios where ranges and limits are crucial. From financial modeling to engineering, the ability to work with inequalities provides a more nuanced and realistic way to represent mathematical relationships. Keep practicing these techniques, and you'll find yourself becoming increasingly comfortable and proficient in tackling more complex mathematical challenges. Don't hesitate to revisit these examples or try similar problems. The more you practice, the more intuitive solving inequalities will become.

For further exploration and resources on inequalities and rational numbers, you can visit Khan Academy's section on inequalities, a fantastic resource for learning and reinforcing mathematical concepts. They offer comprehensive explanations, practice exercises, and video tutorials that can further deepen your understanding.