Example Question - isolating absolute value
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Solving Absolute Value Inequalities
To solve the inequality \( 9 - 8 | r + 5 | > -11 \), we must isolate the absolute value expression on one side of the inequality. Here are the steps to solve the inequality: 1. Add or subtract quantities from both sides of the inequality to move constant terms away from the absolute value expression: \[ 9 - 8 | r + 5 | + 11 > -11 + 11 \] \[ 20 - 8 | r + 5 | > 0 \] 2. Now we need to isolate the absolute value expression. To do this, we can subtract 20 from both sides: \[ 20 - 8 | r + 5 | - 20 > 0 - 20 \] \[ -8 | r + 5 | > -20 \] 3. To get rid of the coefficient in front of the absolute value, we divide both sides by -8. Remember that when you divide by a negative number, you have to flip the inequality sign: \[ \frac{-8 | r + 5 |}{-8} < \frac{-20}{-8} \] \[ | r + 5 | < \frac{20}{8} \] 4. Simplify the fraction: \[ | r + 5 | < 2.5 \] This inequality says that the distance from the point \( r + 5 \) to 0 on the number line is less than 2.5. We can break this down into two separate inequalities: If \( r + 5 \) is positive: \[ r + 5 < 2.5 \] If \( r + 5 \) is negative: \[ r + 5 > -2.5 \] Now solve for \( r \) in each inequality: For the case where \( r + 5 \) is positive: \[ r < 2.5 - 5 \] \[ r < -2.5 \] For the case where \( r + 5 \) is negative: \[ r > -2.5 - 5 \] \[ r > -7.5 \] So the solution to the inequality is: \[ -7.5 < r < -2.5 \] This means that \( r \) is greater than -7.5 and less than -2.5.
Solving Absolute Value Inequality
To solve the inequality \( 9 - 8|r + 5| > -11 \), we need to isolate the absolute value expression on one side. First, let's move the constant term on the left to the other side by subtracting 9 from both sides: \( 9 - 8|r + 5| - 9 > -11 - 9 \) \( -8|r + 5| > -20 \) Next, to get \( |r + 5| \) by itself, divide both sides of the inequality by -8. Remember that dividing by a negative number flips the direction of the inequality: \( \frac{-8|r + 5|}{-8} < \frac{-20}{-8} \) \( |r + 5| < \frac{20}{8} \) \( |r + 5| < \frac{5}{2} \) or \( |r + 5| < 2.5 \) The absolute value inequality \( |r + 5| < 2.5 \) means that \( r + 5 \) must be less than 2.5 and greater than -2.5 (because the distance from zero is less than 2.5). So we can break this into two separate inequalities: \( r + 5 < 2.5 \) \( r + 5 > -2.5 \) Let's solve each of these: For \( r + 5 < 2.5 \): Subtract 5 from both sides: \( r < 2.5 - 5 \) \( r < -2.5 \) For \( r + 5 > -2.5 \): Subtract 5 from both sides: \( r > -2.5 - 5 \) \( r > -7.5 \) Combining both inequalities, we get the solution for \( r \): \( -7.5 < r < -2.5 \) This is the set of all real numbers greater than -7.5 and less than -2.5.
Solving Inequalities Involving Absolute Value
To solve the inequality \(-9|c| < -50\), let's start by isolating the absolute value term. Firstly, divide both sides by -9 to get the absolute value by itself. It's important to remember that when we divide or multiply both sides of an inequality by a negative number, the direction of the inequality symbol changes. So the inequality will reverse in this case. \[ -9|c| < -50 \quad / :(-9) \] \[ |c| > \frac{-50}{-9} \] \[ |c| > \frac{50}{9} \] Since the absolute value of a number is always nonnegative, we can write this as two separate inequalities: \[ c > \frac{50}{9} \quad or \quad c < -\frac{50}{9} \] So the solution to the inequality is that \(c\) is either greater than \(\frac{50}{9}\) or less than \(-\frac{50}{9}\).
Absolute Value Inequality Solution
The inequality is \( 3|d| + 5 < 47 \). First, we'll isolate the absolute value expression by subtracting 5 from both sides of the inequality: \( 3|d| + 5 - 5 < 47 - 5 \) \( 3|d| < 42 \) Now, divide both sides by 3 to solve for the absolute value of d: \( \frac{3|d|}{3} < \frac{42}{3} \) \( |d| < 14 \) Since we have an absolute value inequality, we know that \( |d| < 14 \) means that d is less than 14 and greater than -14. Therefore, the solution set for d is: \( -14 < d < 14 \)
Solving Absolute Value Inequality and Writing as a Compound Inequality
To solve the inequality \( 4|v - 3| \leq 28 \) and write it as a compound inequality, follow these steps: 1. First, isolate the absolute value on one side by dividing both sides of the inequality by 4: \[ |v - 3| \leq \frac{28}{4} \] \[ |v - 3| \leq 7 \] 2. Now, remember that the absolute value inequality \( |x| \leq a \) can be written as the compound inequality \( -a \leq x \leq a \). Applying this to your inequality, you get: \[ -7 \leq v - 3 \leq 7 \] 3. Adding 3 to all parts of the compound inequality to solve for \( v \) gives us: \[ -7 + 3 \leq v - 3 + 3 \leq 7 + 3 \] \[ -4 \leq v \leq 10 \] So, the solution to the inequality \( 4|v - 3| \leq 28 \) expressed as a compound inequality is \( -4 \leq v \leq 10 \).
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, ∞)
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