Example Question - absolute value inequalities
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Solving Absolute Value Inequalities with Two Cases
To solve the inequality \(7 \leq |7y - 9|\), we must consider two cases because the absolute value function \(|x|\) is defined as \(x\) when \(x \geq 0\) and \(-x\) when \(x < 0\). **Case 1**: \(7y - 9 \geq 0\) If \(7y - 9\) is non-negative, the inequality is simply: \(7 \leq 7y - 9\) Now, we solve for \(y\): \(7 + 9 \leq 7y\) \(16 \leq 7y\) Divide both sides by 7: \(y \geq \frac{16}{7}\) **Case 2**: \(7y - 9 < 0\) If \(7y - 9\) is negative, then \(|7y - 9| = -(7y - 9)\), so the inequality becomes: \(7 \leq -(7y - 9)\) Now, we solve for \(y\): \(7 \leq -7y + 9\) Rearrange: \(7y \leq 9 - 7\) \(7y \leq 2\) Divide both sides by 7: \(y \leq \frac{2}{7}\) Combining the results from both cases gives us the complete solution set for the inequality: \(\frac{2}{7} \geq y \geq \frac{16}{7}\) However, if you look carefully at the inequalities derived from both cases, you'll notice that there's actually no overlap between the two solution sets, since it's not possible for a number \(y\) to be simultaneously greater than or equal to approximately \(2.29\) and less than or equal to approximately \(0.29\). This suggests we need to correct the combined inequality to reflect the actual solution. The correct combined solution for the inequality considering both cases would be: \(y \leq \frac{2}{7}\) or \(y \geq \frac{16}{7}\) Thus, the solution set for the inequality \(7 \leq |7y - 9|\) is that \(y\) can be in the interval \((-\infty, \frac{2}{7}]\) or \([\frac{16}{7}, +\infty)\).
Solving Absolute Value Inequalities
The inequality presented here is \(7 \leq |7y - 9|\). To solve this, we need to consider the absolute value |7y - 9| and split the inequality into two cases because the expression inside the absolute value can be either positive or negative. The absolute value inequality says that the expression inside the absolute value is either greater than or equal to 7 or less than or equal to -7. First, we'll handle the case where the expression inside the absolute value is non-negative: 1. \(7y - 9 \geq 7\) 2. \(7y \geq 16\) 3. \(y \geq \frac{16}{7}\) Next, we handle the case where the expression inside the absolute value is non-positive: 1. \(7y - 9 \leq -7\) 2. \(7y \leq 2\) 3. \(y \leq \frac{2}{7}\) Combining both inequalities, we get the solution set for y: \(y \leq \frac{2}{7}\) or \(y \geq \frac{16}{7}\). Graphically, this means y is either in the interval \((-\infty, \frac{2}{7}]\) or in the interval \([\frac{16}{7}, \infty)\).
Solving Absolute Value Inequalities
The inequality \( |t - 75| \leq 15 \) represents the values of \( t \) that are at most 15 units away from 75 on the number line, either less than or equal to 75 or greater than or equal to 75, but not further away than 15 units in either direction. To solve this absolute value inequality, we'll break it into two separate inequalities: 1. \( t - 75 \leq 15 \) 2. \( - (t - 75) \leq 15 \) which is the same as \( t - 75 \geq -15 \) Now, we solve each inequality: 1. \( t \leq 15 + 75 \) \( t \leq 90 \) 2. \( t - 75 \geq -15 \) \( t \geq -15 + 75 \) \( t \geq 60 \) Combining these two inequalities, we get: \[ 60 \leq t \leq 90 \] Therefore, the solution to the absolute value inequality is all real numbers \( t \) between 60 and 90, inclusive.
Solving Absolute Value Inequalities with a Non-Negative Constant
This inequality involves an absolute value. The statement |x| ≥ k, where k is a non-negative number, is equivalent to saying that x ≤ -k or x ≥ k, since the absolute value indicates the distance of a number from zero, not the direction. Given the inequality: 50 ≤ |1 - g + 950| we can split this into two separate inequalities because the expression inside the absolute value must either be greater than or equal to 50 or less than or equal to -50. Therefore, we have: 1. 1 - g + 950 ≥ 50 2. 1 - g + 950 ≤ -50 Let's solve each inequality. For the first inequality: 1 - g + 950 ≥ 50 Subtract 950 from both sides: 1 - g ≥ 50 - 950 1 - g ≥ -900 Now, add g to both sides: 1 ≥ g - 900 Finally, add 900 to both sides: g ≤ 901 For the second inequality: 1 - g + 950 ≤ -50 Subtract 950 from both sides: 1 - g ≤ -50 - 950 1 - g ≤ -1000 Add g to both sides: 1 + g ≤ -1000 Add 1 to both sides: g ≤ -1001 Combining both sets of g values, we have: g ≤ 901 and g ≤ -1001 However, since g cannot be simultaneously less than or equal to both 901 and -1001, we consider the entire range of g that satisfies either condition. Thus, the solution to the inequality is: g ≤ 901 Since -1001 is much less than 901, the latter encompasses all possible values of g that could satisfy the inequality.
Solving Absolute Value Inequalities
To solve the inequality \( 8 - 7|1 - 6s| < -6 \), follow these steps: 1. First, isolate the absolute value term on one side of the inequality: \( 8 - 7|1 - 6s| + 6 < 0 \) \( 14 - 7|1 - 6s| < 0 \) \( -7|1 - 6s| < -14 \) Now divide by -7, remembering that dividing by a negative number will flip the inequality sign: \( |1 - 6s| > 2 \) 2. Next, we need to consider the two cases due to the absolute value, where \(1 - 6s\) can be either greater than 2 or less than -2. **Case 1 - Positive case:** \( 1 - 6s > 2 \) Subtract 1 from both sides: \( -6s > 1 \) Divide by -6 and flip the inequality sign: \( s < \frac{1}{6} \) **Case 2 - Negative case:** \( 1 - 6s < -2 \) Subtract 1 from both sides: \( -6s < -3 \) Divide by -6 and flip the inequality sign: \( s > \frac{1}{2} \) 3. Combine both cases to get the final solution: \( s < \frac{1}{6} \) or \( s > \frac{1}{2} \) The solution to the inequality is all \( s \) that satisfy either \( s < \frac{1}{6} \) or \( s > \frac{1}{2} \).
Solving Absolute Value Inequalities
Sure, let's solve the inequality presented in the image: \[ |2p| - 8 \geq 13 \] To solve this, you first need to isolate the absolute value expression on one side of the inequality. Let's move the -8 to the other side by adding 8 to both sides: \[ |2p| \geq 13 + 8 \] \[ |2p| \geq 21 \] Now that the absolute value is isolated, we can split the inequality into two separate cases, because the absolute value of a number is either equal to the number itself if the number is positive or equal to the negative of the number if it is negative. So we have: \[ 2p \geq 21 \quad \text{or} \quad 2p \leq -21 \] Now, solve each inequality for \( p \): For the first inequality: \[ 2p \geq 21 \] \[ p \geq \frac{21}{2} \] \[ p \geq 10.5 \] For the second inequality: \[ 2p \leq -21 \] \[ p \leq \frac{-21}{2} \] \[ p \leq -10.5 \] Therefore, the solution to the inequality is \( p \leq -10.5 \) or \( p \geq 10.5 \). This means \( p \) can be any number less than or equal to -10.5 or any number greater than or equal to 10.5.
Solving Absolute Value Inequalities
Certainly! We have an inequality involving an absolute value: |u - 7| > 29. Recall that |x| > a, where a is a positive number, implies two situations: 1. x > a 2. x < -a Applying this to our inequality, we split it into two separate inequalities: u - 7 > 29 and u - 7 < -29 For the first inequality (u - 7 > 29), we solve for u by adding 7 to both sides: u > 29 + 7 u > 36 For the second inequality (u - 7 < -29), we again solve for u by adding 7 to both sides: u < -29 + 7 u < -22 Therefore, the solutions to the inequality |u - 7| > 29 are: u > 36 or u < -22 These inequalities represent the range of values that u can take to satisfy the original inequality.
Solving Absolute Value Inequalities
To solve the inequality \( |6 - w| < 59 \), we approach it by considering the definition of the absolute value. The absolute value \( |a| \) of a number is its distance from zero on the number line, regardless of direction. This means \( |6 - w| < 59 \) tells us that the expression \( 6 - w \) must be less than 59 units away from 0 on the number line. We must consider two cases due to the absolute value: 1. \( 6 - w \) is positive or zero: \[ 6 - w < 59 \] \[ -w < 59 - 6 \] \[ -w < 53 \] \[ w > -53 \] (When we divide or multiply by a negative number, we reverse the inequality sign). 2. \( 6 - w \) is negative: \[ -(6 - w) < 59 \] \[ w - 6 < 59 \] \[ w < 59 + 6 \] \[ w < 65 \] Therefore, combining both cases, the solution to the inequality is: \[ -53 < w < 65 \] This tells us w must be greater than -53 and less than 65.
Solving Absolute Value Inequalities
To solve the inequality |6 - w| < 59, you will need to consider two cases due to the absolute value. Case 1: When 6 - w is positive or zero, 6 - w < 59 Now, solve for w: -w < 59 - 6 -w < 53 Multiply both sides by -1 (and remember to reverse the inequality when you multiply or divide by a negative number): w > -53 Case 2: When 6 - w is negative, -(6 - w) < 59 -w + 6 < 59 Now, solve for w: -w < 59 - 6 -w < 53 Again, multiply both sides by -1 (and reverse the inequality): w > -53 Since both cases lead to the same inequality, w > -53, the solution to the inequality |6 - w| < 59 is: w > -53 This means that w can be any number greater than -53.
Solving Absolute Value Inequalities
To solve the absolute value inequality \( |4q - 1| \leq 85 \), we must consider the two cases of the absolute value definition: 1. The expression inside the absolute value, \( 4q - 1 \), is non-negative. 2. The expression inside the absolute value, \( 4q - 1 \), is negative. For the first case, when \( 4q - 1 \) is non-negative: \[ 4q - 1 \leq 85 \] To solve this inequality, add 1 to both sides: \[ 4q \leq 86 \] Now, divide both sides by 4: \[ q \leq 21.5 \] For the second case, when \( 4q - 1 \) is negative, we consider the opposite of the expression, which gives us: \[ -(4q - 1) \leq 85 \] Simplify the left side by distributing the negative sign: \[ -4q + 1 \leq 85 \] Now, to isolate the q term, subtract 1 from both sides: \[ -4q \leq 84 \] Divide both sides by -4 and remember to reverse the inequality sign because we are dividing by a negative number: \[ q \geq -21 \] Putting both cases together gives us the compound inequality: \[ -21 \leq q \leq 21.5 \] This is the solution set for the inequality \( |4q - 1| \leq 85 \).
Solving Absolute Value Inequalities
To solve the inequality |4q - 1| ≤ 85, you need to consider the definition of the absolute value. The absolute value of a number is the distance of that number from 0 on the number line, which is always nonnegative. Therefore, |x| ≤ a means that x is within the distance of a from 0. This leads to two scenarios: x ≤ a and x ≥ -a. Applying this idea to |4q - 1| ≤ 85, we get two inequalities: 1. 4q - 1 ≤ 85 2. 4q - 1 ≥ -85 Now, let's solve each inequality separately. For the first inequality: 4q - 1 ≤ 85 Add 1 to both sides: 4q ≤ 86 Divide both sides by 4: q ≤ 86 / 4 q ≤ 21.5 For the second inequality: 4q - 1 ≥ -85 Add 1 to both sides: 4q ≥ -84 Divide both sides by 4: q ≥ -84 / 4 q ≥ -21 Combining these two inequalities gives us the compound inequality: -21 ≤ q ≤ 21.5 This is the solution to the original inequality, which means that q must be greater than or equal to -21 and less than or equal to 21.5.
Solving Absolute Value Inequalities
To solve the inequality \( |c| - 1 \geq 11 \), let's first isolate the absolute value expression on one side. \( |c| - 1 \geq 11 \) Add 1 to both sides to get: \( |c| \geq 12 \) This inequality means that the distance of \( c \) from 0 on the number line is greater than or equal to 12. Absolute value inequalities like this one split into two cases because the variable inside the absolute value can be either positive or negative. Case 1: If \( c \) is positive or zero, then we remove the absolute value: \( c \geq 12 \) Case 2: If \( c \) is negative, the absolute value makes it positive, so we have: \( -c \geq 12 \) Multiply both sides by -1, remembering to reverse the inequality sign since multiplying an inequality by a negative number reverses the direction of the inequality: \( c \leq -12 \) Now, we combine both cases into a compound inequality: \( c \leq -12 \) or \( c \geq 12 \) These inequalities represent all the possible values of \( c \) that make the original inequality true. In interval notation, this solution would be: \( (-\infty, -12] \cup [12, \infty) \)
Solving Absolute Value Inequalities
To solve the inequality \( |c| - 1 \geq 11 \), you need to isolate the absolute value on one side of the inequality. First, add 1 to both sides of the inequality: \[ |c| - 1 + 1 \geq 11 + 1 \] \[ |c| \geq 12 \] The absolute value \( |c| \) represents the distance from 0 on the number line, and it can be either positive or negative. Since \( |c| \geq 12 \), this means that c can either be greater than or equal to 12, or less than or equal to -12. Split the inequality into two cases: 1. \( c \geq 12 \) (when c is positive or zero) 2. \( c \leq -12 \) (when c is negative) Therefore, the solution set for \( |c| - 1 \geq 11 \) is \( c \geq 12 \) or \( c \leq -12 \).
Solving Absolute Value Inequalities
To solve the inequality |c| - 1 ≥ 11, follow these steps: 1. Start by isolating the absolute value on one side of the inequality: |c| - 1 + 1 ≥ 11 + 1 |c| ≥ 12 2. Next, understand that an absolute value being greater than or equal to a positive number means the expression inside the absolute value can be either greater than or equal to that number or less than or equal to the negative of that number. In other words, if |a| ≥ b and b > 0, then a ≥ b or a ≤ -b. So for |c| ≥ 12, we have two cases: Case 1: c ≥ 12 Case 2: c ≤ -12 The solution to the inequality |c| - 1 ≥ 11 is therefore all c that are greater than or equal to 12, combined with all c that are less than or equal to -12. This can be expressed as: c ≥ 12 or c ≤ -12 Or, in interval notation: c ∈ (-∞, -12] ∪ [12, ∞)
Solving Absolute Value Inequalities
To solve the inequality \( |8r| + 2 > 42 \), we will first isolate the absolute value part by subtracting 2 from both sides of the inequality: \[ |8r| + 2 - 2 > 42 - 2 \] \[ |8r| > 40 \] Now, by the definition of absolute value, \( |8r| > 40 \) means that 8r is either greater than 40 or less than -40. We can split this into two separate inequalities to represent the compound inequality: 1) \( 8r > 40 \) 2) \( 8r < -40 \) Next, we can solve each inequality for r. Starting with the first inequality: \[ 8r > 40 \] \[ r > \frac{40}{8} \] \[ r > 5 \] For the second inequality: \[ 8r < -40 \] \[ r < \frac{-40}{8} \] \[ r < -5 \] Now, we present these two solutions as a compound inequality: \[ r < -5 \text{ or } r > 5 \] This is the solution to the original inequality \( |8r| + 2 > 42 \).
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