Solving Inequalities And Equations With Fractions And Absolute Values

In this comprehensive guide, we delve into solving various inequalities and equations involving fractions and absolute values. These types of problems often require a combination of algebraic manipulation, careful consideration of domain restrictions, and a solid understanding of the properties of inequalities and absolute values. We will tackle each problem step-by-step, providing detailed explanations and strategies to help you master these concepts. The examples cover a range of complexities, offering a thorough exploration of the techniques needed to solve them effectively. Whether you're a student looking to improve your algebra skills or someone seeking a refresher, this article will equip you with the tools and knowledge to confidently approach these mathematical challenges.

14. Solving the Inequality (2x+3)/x < 2

To address the inequality (2x+3)/x < 2, our initial step is to rearrange the inequality to have zero on one side. This allows us to analyze the sign of the expression more effectively. We accomplish this by subtracting 2 from both sides of the inequality, leading us to the expression (2x+3)/x - 2 < 0. The next crucial step involves finding a common denominator to combine the terms. In this case, the common denominator is x. By rewriting 2 as 2x/x, we can combine the fractions into a single expression. This gives us (2x+3 - 2x) / x < 0. Simplifying the numerator, we arrive at 3/x < 0. Now, the problem is significantly simplified. We need to determine the values of x for which the fraction 3/x is negative.

Since the numerator is a positive constant (3), the fraction will be negative only when the denominator, x, is negative. Therefore, the solution to the inequality is x < 0. However, it is essential to consider the domain of the original expression. The original inequality (2x+3)/x < 2 contains a fraction with x in the denominator. This implies that x cannot be equal to zero because division by zero is undefined. Thus, we must exclude x = 0 from our solution set. Combining these considerations, the solution to the inequality is all x values less than 0, which can be written as x < 0. This is the final solution to the given inequality.

In summary, we started with the inequality (2x+3)/x < 2, manipulated it algebraically to the form 3/x < 0, and determined that the solution is x < 0. It is crucial to remember to check the original inequality for any domain restrictions, which in this case is x ≠ 0. This step-by-step approach ensures we arrive at the correct solution while adhering to the mathematical rules and principles.

15. Solving the Inequality (x^2-1)/(x-2) ≤ 0

To solve the inequality (x^2-1)/(x-2) ≤ 0, we first need to factor the numerator. The numerator, x^2 - 1, is a difference of squares, which can be factored into (x - 1)(x + 1). Thus, the inequality becomes ((x - 1)(x + 1))/(x - 2) ≤ 0. Next, we identify the critical points by setting each factor in the numerator and the denominator equal to zero. These critical points are the values of x that make the expression equal to zero or undefined. Setting x - 1 = 0 gives us x = 1. Setting x + 1 = 0 gives us x = -1. Setting x - 2 = 0 gives us x = 2. These critical points divide the number line into intervals, and the sign of the expression may change at these points. The critical points are x = -1, 1, and 2.

Now, we create a sign chart to analyze the sign of the expression in each interval. The intervals are (-∞, -1), (-1, 1), (1, 2), and (2, ∞). We choose test values within each interval to determine the sign of each factor and the overall expression. For the interval (-∞, -1), we can choose x = -2. Plugging this into the factors, we get (-2 - 1) = -3, (-2 + 1) = -1, and (-2 - 2) = -4. The expression becomes ((-3)(-1))/(-4) = 3/(-4), which is negative. For the interval (-1, 1), we can choose x = 0. The factors become (0 - 1) = -1, (0 + 1) = 1, and (0 - 2) = -2. The expression becomes ((-1)(1))/(-2) = -1/(-2), which is positive. For the interval (1, 2), we can choose x = 1.5. The factors become (1.5 - 1) = 0.5, (1.5 + 1) = 2.5, and (1.5 - 2) = -0.5. The expression becomes (0.5 * 2.5)/(-0.5), which is negative. For the interval (2, ∞), we can choose x = 3. The factors become (3 - 1) = 2, (3 + 1) = 4, and (3 - 2) = 1. The expression becomes (2 * 4)/1, which is positive.

From the sign chart, we see that the expression is less than or equal to zero in the intervals (-∞, -1] and [1, 2). We include x = -1 and x = 1 because the inequality is non-strict (≤). However, we exclude x = 2 because the expression is undefined at this point due to division by zero. Therefore, the solution to the inequality is x ∈ (-∞, -1] ∪ [1, 2).

16. Solving the Inequality (x^2-4x+3)/(x+1) > 0

To solve the inequality (x^2-4x+3)/(x+1) > 0, we begin by factoring the numerator. The quadratic x^2 - 4x + 3 can be factored into (x - 3)(x - 1). This transforms the inequality into ((x - 3)(x - 1))/(x + 1) > 0. Next, we identify the critical points, which are the values of x that make either the numerator or the denominator equal to zero. These critical points help us determine the intervals where the expression may change its sign. Setting each factor to zero, we have x - 3 = 0, which gives x = 3; x - 1 = 0, which gives x = 1; and x + 1 = 0, which gives x = -1. Thus, the critical points are x = -1, 1, and 3.

Now, we construct a sign chart using these critical points to analyze the sign of the expression in the intervals they create. The intervals are (-∞, -1), (-1, 1), (1, 3), and (3, ∞). We select test values within each interval to determine the sign of each factor and, consequently, the sign of the entire expression. For the interval (-∞, -1), let's choose x = -2. Plugging this value into the factors, we have (-2 - 3) = -5, (-2 - 1) = -3, and (-2 + 1) = -1. The expression becomes ((-5)(-3))/(-1) = 15/(-1), which is negative. For the interval (-1, 1), we can choose x = 0. The factors become (0 - 3) = -3, (0 - 1) = -1, and (0 + 1) = 1. The expression becomes ((-3)(-1))/1 = 3/1, which is positive. For the interval (1, 3), let's choose x = 2. The factors become (2 - 3) = -1, (2 - 1) = 1, and (2 + 1) = 3. The expression becomes ((-1)(1))/3 = -1/3, which is negative. Lastly, for the interval (3, ∞), we choose x = 4. The factors become (4 - 3) = 1, (4 - 1) = 3, and (4 + 1) = 5. The expression becomes (1 * 3)/5 = 3/5, which is positive.

From the sign chart analysis, we observe that the expression is greater than zero in the intervals (-1, 1) and (3, ∞). Therefore, the solution to the inequality is x ∈ (-1, 1) ∪ (3, ∞). Note that we use open intervals because the original inequality is strict (>), meaning the critical points themselves are not included in the solution set.

17. Solving the Inequality (-2x^2+13x-15)/(x2+x+1) < 0

To solve the inequality (-2x^2+13x-15)/(x2+x+1) < 0, we first factor the numerator. The quadratic expression -2x^2 + 13x - 15 can be factored as -(2x - 3)(x - 5). Therefore, the inequality becomes (-(2x - 3)(x - 5))/(x^2 + x + 1) < 0. Next, we need to analyze the denominator, x^2 + x + 1. To determine if it can be factored, we calculate the discriminant, which is given by the formula Δ = b^2 - 4ac. In this case, a = 1, b = 1, and c = 1. Thus, Δ = 1^2 - 4(1)(1) = 1 - 4 = -3. Since the discriminant is negative, the quadratic x^2 + x + 1 has no real roots and cannot be factored over the real numbers. Moreover, since the leading coefficient is positive, this quadratic is always positive for all real values of x. This simplifies our analysis significantly because the denominator will not change the sign of the expression.

Now, we focus on the numerator, -(2x - 3)(x - 5). The critical points are found by setting each factor to zero. From 2x - 3 = 0, we get x = 3/2, and from x - 5 = 0, we get x = 5. Thus, the critical points are x = 3/2 and x = 5. These points divide the number line into three intervals: (-∞, 3/2), (3/2, 5), and (5, ∞). We create a sign chart to analyze the sign of the expression in each interval.

We choose test values within each interval. For (-∞, 3/2), let's choose x = 0. The factors become -(2(0) - 3) = 3 and (0 - 5) = -5. The expression is (3)(-5), which is negative. Since the denominator is always positive, the entire expression is negative in this interval. For (3/2, 5), we can choose x = 2. The factors become -(2(2) - 3) = -1 and (2 - 5) = -3. The expression is (-1)(-3), which is positive. Again, since the denominator is always positive, the entire expression is positive in this interval. For (5, ∞), let's choose x = 6. The factors become -(2(6) - 3) = -9 and (6 - 5) = 1. The expression is (-9)(1), which is negative. The entire expression is negative in this interval.

Since we want to find where the expression is less than zero, we consider the intervals where the sign is negative. Thus, the solution to the inequality is x ∈ (-∞, 3/2) ∪ (5, ∞). The critical points are not included because the original inequality is strict (<).

18. Solving the Inequality (2x-3)/x < -x

To solve the inequality (2x-3)/x < -x, the first step is to rearrange the inequality so that we have zero on one side. This allows us to combine the terms and analyze the resulting expression more easily. We add x to both sides of the inequality, which gives us (2x-3)/x + x < 0. Now, we need to combine the terms on the left side. To do this, we find a common denominator, which in this case is x. We rewrite x as x^2/x, and the inequality becomes (2x-3)/x + x^2/x < 0. Combining the fractions, we get (2x - 3 + x^2)/x < 0. Rearranging the terms in the numerator, we have (x^2 + 2x - 3)/x < 0.

Next, we factor the quadratic expression in the numerator. The quadratic x^2 + 2x - 3 can be factored into (x + 3)(x - 1). This gives us the inequality ((x + 3)(x - 1))/x < 0. Now, we identify the critical points by setting each factor in the numerator and the denominator equal to zero. These critical points are the values of x that make the expression equal to zero or undefined. Setting x + 3 = 0 gives us x = -3. Setting x - 1 = 0 gives us x = 1. Setting x = 0 (from the denominator) gives us x = 0. Thus, the critical points are x = -3, 0, and 1.

We use these critical points to create a sign chart and analyze the sign of the expression in each interval. The intervals are (-∞, -3), (-3, 0), (0, 1), and (1, ∞). For the interval (-∞, -3), let's choose x = -4. The factors become (-4 + 3) = -1, (-4 - 1) = -5, and (-4). The expression becomes ((-1)(-5))/(-4) = 5/(-4), which is negative. For the interval (-3, 0), we can choose x = -1. The factors become (-1 + 3) = 2, (-1 - 1) = -2, and (-1). The expression becomes (2(-2))/(-1) = -4/(-1), which is positive. For the interval (0, 1), let's choose x = 0.5. The factors become (0.5 + 3) = 3.5, (0.5 - 1) = -0.5, and (0.5). The expression becomes (3.5(-0.5))/(0.5), which is negative. For the interval (1, ∞), we choose x = 2. The factors become (2 + 3) = 5, (2 - 1) = 1, and (2). The expression becomes (5 * 1)/2, which is positive.

From the sign chart, we see that the expression is less than zero in the intervals (-∞, -3) and (0, 1). Therefore, the solution to the inequality is x ∈ (-∞, -3) ∪ (0, 1). Note that we use open intervals because the original inequality is strict (<), and the value x = 0 is excluded since it makes the denominator zero.

19. Solving the Inequality (2x-3)/x - 1 < -x + 3

To solve the inequality (2x-3)/x - 1 < -x + 3, we first want to rearrange the inequality so that all terms are on one side, and we have zero on the other side. This helps in simplifying and analyzing the inequality. We add x and subtract 3 from both sides, resulting in (2x-3)/x - 1 + x - 3 < 0. Next, we combine the terms on the left side. First, let's combine the constant terms: -1 - 3 = -4. So the inequality becomes (2x-3)/x + x - 4 < 0. To combine all terms into a single fraction, we need a common denominator, which in this case is x. We rewrite x as x^2/x and -4 as -4x/x. The inequality then becomes (2x-3)/x + x^2/x - 4x/x < 0. Combining the fractions, we get (2x - 3 + x^2 - 4x)/x < 0. Simplifying the numerator, we have (x^2 - 2x - 3)/x < 0.

Now, we factor the quadratic expression in the numerator. The quadratic x^2 - 2x - 3 can be factored into (x - 3)(x + 1). This gives us the inequality ((x - 3)(x + 1))/x < 0. Next, we identify the critical points by setting each factor in the numerator and the denominator equal to zero. These critical points are the values of x that make the expression equal to zero or undefined. Setting x - 3 = 0 gives us x = 3. Setting x + 1 = 0 gives us x = -1. Setting x = 0 (from the denominator) gives us x = 0. Thus, the critical points are x = -1, 0, and 3.

We use these critical points to create a sign chart and analyze the sign of the expression in each interval. The intervals are (-∞, -1), (-1, 0), (0, 3), and (3, ∞). For the interval (-∞, -1), let's choose x = -2. The factors become (-2 - 3) = -5, (-2 + 1) = -1, and (-2). The expression becomes ((-5)(-1))/(-2) = 5/(-2), which is negative. For the interval (-1, 0), we can choose x = -0.5. The factors become (-0.5 - 3) = -3.5, (-0.5 + 1) = 0.5, and (-0.5). The expression becomes ((-3.5)(0.5))/(-0.5), which is positive. For the interval (0, 3), let's choose x = 1. The factors become (1 - 3) = -2, (1 + 1) = 2, and (1). The expression becomes ((-2)(2))/(1), which is negative. For the interval (3, ∞), we choose x = 4. The factors become (4 - 3) = 1, (4 + 1) = 5, and (4). The expression becomes (1 * 5)/4, which is positive.

From the sign chart, we see that the expression is less than zero in the intervals (-∞, -1) and (0, 3). Therefore, the solution to the inequality is x ∈ (-∞, -1) ∪ (0, 3). We use open intervals because the original inequality is strict (<), and the value x = 0 is excluded since it makes the denominator zero.

20. Solving the Absolute Value Equation |2-3x| = x^2 + 2

To solve the absolute value equation |2-3x| = x^2 + 2, we need to consider two cases based on the definition of absolute value. The absolute value of an expression is the distance of that expression from zero, which means it can be either the expression itself (if it's non-negative) or the negation of the expression (if it's negative).

Case 1: 2 - 3x ≥ 0. In this case, |2 - 3x| = 2 - 3x. So, the equation becomes 2 - 3x = x^2 + 2. Rearranging the terms to form a quadratic equation, we have x^2 + 3x = 0. We can factor out an x from the left side, giving us x(x + 3) = 0. This yields two potential solutions: x = 0 and x = -3. We must check if these solutions satisfy the condition 2 - 3x ≥ 0. For x = 0, 2 - 3(0) = 2, which is greater than or equal to 0, so x = 0 is a valid solution. For x = -3, 2 - 3(-3) = 2 + 9 = 11, which is also greater than or equal to 0, so x = -3 is also a valid solution.

Case 2: 2 - 3x < 0. In this case, |2 - 3x| = -(2 - 3x) = 3x - 2. So, the equation becomes 3x - 2 = x^2 + 2. Rearranging the terms to form a quadratic equation, we have x^2 - 3x + 4 = 0. To determine if this quadratic equation has real solutions, we calculate the discriminant, which is given by the formula Δ = b^2 - 4ac. Here, a = 1, b = -3, and c = 4. Thus, Δ = (-3)^2 - 4(1)(4) = 9 - 16 = -7. Since the discriminant is negative, there are no real solutions for this quadratic equation in this case.

Combining the results from both cases, the real solutions to the absolute value equation |2-3x| = x^2 + 2 are x = 0 and x = -3. These are the only values of x that satisfy the given equation.

21. Solving the Absolute Value Equation |2-3x| = x^2

To solve the absolute value equation |2-3x| = x^2, we must consider two separate cases arising from the definition of absolute value. The expression inside the absolute value, 2-3x, can be either positive or negative, and we need to address both possibilities to find all solutions.

Case 1: 2 - 3x ≥ 0. In this case, the absolute value |2 - 3x| is equal to 2 - 3x itself. So, the equation becomes 2 - 3x = x^2. Rearranging the terms to form a standard quadratic equation, we get x^2 + 3x - 2 = 0. To solve this quadratic equation, we can use the quadratic formula, which is given by x = (-b ± √(b^2 - 4ac)) / (2a). In this equation, a = 1, b = 3, and c = -2. Plugging these values into the quadratic formula, we get x = (-3 ± √(3^2 - 4(1)(-2))) / (2(1)) = (-3 ± √(9 + 8)) / 2 = (-3 ± √17) / 2. So, we have two potential solutions: x = (-3 + √17) / 2 and x = (-3 - √17) / 2. Now, we need to check if these solutions satisfy the condition 2 - 3x ≥ 0. For x = (-3 + √17) / 2, 2 - 3((-3 + √17) / 2) = 2 + (9 - 3√17) / 2 = (4 + 9 - 3√17) / 2 = (13 - 3√17) / 2. Since √17 is approximately 4.12, 3√17 is approximately 12.36, so 13 - 3√17 is positive, and thus this solution satisfies the condition. For x = (-3 - √17) / 2, 2 - 3((-3 - √17) / 2) = 2 + (9 + 3√17) / 2 = (4 + 9 + 3√17) / 2 = (13 + 3√17) / 2, which is clearly positive, so this solution also satisfies the condition.

Case 2: 2 - 3x < 0. In this case, the absolute value |2 - 3x| is equal to the negation of 2 - 3x, which is 3x - 2. So, the equation becomes 3x - 2 = x^2. Rearranging the terms to form a standard quadratic equation, we get x^2 - 3x + 2 = 0. This quadratic equation can be factored as (x - 1)(x - 2) = 0. This yields two potential solutions: x = 1 and x = 2. Now, we need to check if these solutions satisfy the condition 2 - 3x < 0. For x = 1, 2 - 3(1) = -1, which is less than 0, so x = 1 is a valid solution. For x = 2, 2 - 3(2) = -4, which is also less than 0, so x = 2 is also a valid solution.

Combining the results from both cases, the solutions to the absolute value equation |2-3x| = x^2 are x = (-3 + √17) / 2, x = (-3 - √17) / 2, x = 1, and x = 2. These are all the values of x that satisfy the given equation.

22. Solving the Absolute Value Inequality |2x^2+3x-2| > 3

To solve the absolute value inequality |2x^2+3x-2| > 3, we need to consider two cases based on the definition of absolute value. The absolute value of an expression is greater than 3 if the expression itself is greater than 3 or if the expression is less than -3.

Case 1: 2x^2 + 3x - 2 > 3. In this case, we have the inequality 2x^2 + 3x - 2 > 3. Subtracting 3 from both sides, we get 2x^2 + 3x - 5 > 0. We can factor the quadratic expression as (2x + 5)(x - 1) > 0. The critical points are found by setting each factor to zero: 2x + 5 = 0 gives x = -5/2, and x - 1 = 0 gives x = 1. These critical points divide the number line into three intervals: (-∞, -5/2), (-5/2, 1), and (1, ∞). We test a value from each interval in the inequality (2x + 5)(x - 1) > 0. For the interval (-∞, -5/2), we can choose x = -3. Plugging this in, we get (2(-3) + 5)(-3 - 1) = (-1)(-4) = 4, which is greater than 0. For the interval (-5/2, 1), we can choose x = 0. Plugging this in, we get (2(0) + 5)(0 - 1) = (5)(-1) = -5, which is not greater than 0. For the interval (1, ∞), we can choose x = 2. Plugging this in, we get (2(2) + 5)(2 - 1) = (9)(1) = 9, which is greater than 0. Thus, the solution for this case is x ∈ (-∞, -5/2) ∪ (1, ∞).

Case 2: 2x^2 + 3x - 2 < -3. In this case, we have the inequality 2x^2 + 3x - 2 < -3. Adding 3 to both sides, we get 2x^2 + 3x + 1 < 0. We can factor the quadratic expression as (2x + 1)(x + 1) < 0. The critical points are found by setting each factor to zero: 2x + 1 = 0 gives x = -1/2, and x + 1 = 0 gives x = -1. These critical points divide the number line into three intervals: (-∞, -1), (-1, -1/2), and (-1/2, ∞). We test a value from each interval in the inequality (2x + 1)(x + 1) < 0. For the interval (-∞, -1), we can choose x = -2. Plugging this in, we get (2(-2) + 1)(-2 + 1) = (-3)(-1) = 3, which is not less than 0. For the interval (-1, -1/2), we can choose x = -3/4. Plugging this in, we get (2(-3/4) + 1)(-3/4 + 1) = (-1/2)(1/4) = -1/8, which is less than 0. For the interval (-1/2, ∞), we can choose x = 0. Plugging this in, we get (2(0) + 1)(0 + 1) = (1)(1) = 1, which is not less than 0. Thus, the solution for this case is x ∈ (-1, -1/2).

Combining the results from both cases, the solution to the absolute value inequality |2x^2+3x-2| > 3 is x ∈ (-∞, -5/2) ∪ (-1, -1/2) ∪ (1, ∞). This is the union of the solutions from the two cases.

23. Solving the Absolute Value Inequality |2-|x-1|| ≤ 1

To solve the absolute value inequality |2-|x-1|| ≤ 1, we need to break it down step by step, considering the different layers of absolute values. The inequality states that the absolute value of the expression 2 - |x - 1| is less than or equal to 1. This means that the expression 2 - |x - 1| must lie between -1 and 1, inclusive. Therefore, we can write the compound inequality as -1 ≤ 2 - |x - 1| ≤ 1.

First, let's isolate the absolute value term. Subtracting 2 from all parts of the inequality, we get -3 ≤ -|x - 1| ≤ -1. Now, we multiply all parts of the inequality by -1, which reverses the inequality signs: 1 ≤ |x - 1| ≤ 3. This compound inequality tells us that the absolute value of x - 1 is between 1 and 3, inclusive. This means we have two separate inequalities to consider: |x - 1| ≥ 1 and |x - 1| ≤ 3.

Let's solve |x - 1| ≥ 1 first. This inequality means that x - 1 is either greater than or equal to 1 or less than or equal to -1. So, we have two cases:

Case 1: x - 1 ≥ 1. Adding 1 to both sides, we get x ≥ 2. Case 2: x - 1 ≤ -1. Adding 1 to both sides, we get x ≤ 0.

So, the solution to |x - 1| ≥ 1 is x ∈ (-∞, 0] ∪ [2, ∞).

Now, let's solve |x - 1| ≤ 3. This inequality means that x - 1 is between -3 and 3, inclusive. So, we can write the compound inequality as -3 ≤ x - 1 ≤ 3. Adding 1 to all parts of the inequality, we get -2 ≤ x ≤ 4. So, the solution to |x - 1| ≤ 3 is x ∈ [-2, 4].

Finally, to find the solution to the original inequality |2-|x-1|| ≤ 1, we need to find the intersection of the solutions to |x - 1| ≥ 1 and |x - 1| ≤ 3. This means we need to find the values of x that satisfy both x ∈ (-∞, 0] ∪ [2, ∞) and x ∈ [-2, 4]. The intersection of these intervals is the set of values that are in both intervals. Looking at the intervals, we see that the intersection is x ∈ [-2, 0] ∪ [2, 4]. This is the final solution to the given absolute value inequality.

In this article, we have explored a variety of problems involving inequalities and equations with fractions and absolute values. We've demonstrated how to approach each type of problem, emphasizing the importance of algebraic manipulation, factoring, sign charts, and considering domain restrictions. By understanding the underlying principles and practicing these techniques, you can confidently tackle even the most challenging problems in this area. Remember to always check your solutions and consider the context of the problem to ensure your answers are accurate and meaningful.