Example Question - solving absolute value inequalities
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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 Inequalities
To solve the inequality -9|c| < -50, divide both sides of the inequality by -9. However, remember that when you divide or multiply an inequality by a negative number, the direction of the inequality sign must be reversed. Dividing both sides by -9, we get: |c| > 50 / 9 The absolute value inequality |c| > 50/9 indicates that c can be greater than 50/9 or less than -50/9. Hence, the solution to the inequality is: c < -50/9 or c > 50/9 Since 50/9 is approximately 5.56, the solution in decimal form is: c < -5.56 or c > 5.56
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
To solve the inequality \( |u - 7| > 29 \), we need to split it into two separate inequalities because the absolute value expression can represent a positive or negative result: 1. If the expression inside the absolute value, \( u - 7 \), is positive, then \( u - 7 > 29 \). To solve for \( u \), add 7 to both sides of the inequality: \( u - 7 + 7 > 29 + 7 \) \( u > 36 \) 2. If the expression inside the absolute value, \( u - 7 \), is negative, then \( -(u - 7) > 29 \). This is equivalent to \( 7 - u > 29 \). To solve for \( u \), first subtract 7 from both sides: \( 7 - u - 7 > 29 - 7 \) \( -u > 22 \) Next, multiply both sides by -1 to isolate \( u \), and remember to reverse the inequality sign when multiplying or dividing by a negative number: \( -u(-1) < 22(-1) \) \( u < -22 \) Combine both solutions to express the full solution to the original inequality: \( u > 36 \) or \( u < -22 \) Hence, \( u \) is any number greater than 36 or any number less than -22.
Solving Inequalities Involving Absolute Value
This is an inequality involving the absolute value of a variable \( d \). To solve for \( d \), we need to consider two cases because the absolute value of a number represents its distance from 0 on the number line, regardless of direction. The inequality is: \[ 3|d| + 5 < 47 \] First, we will isolate the absolute value on one side by subtracting 5 from both sides: \[ 3|d| < 42 \] Next, we will divide both sides by 3 to solve for the absolute value of \( d \): \[ |d| < 14 \] Now let's consider the two cases for \( d \): 1. If \( d \) is non-negative (\( d \geq 0 \)), the absolute value sign can be removed without changing the sign of \( d \): \[ d < 14 \] 2. If \( d \) is negative (\( d < 0 \)), we remove the absolute value sign and reverse the inequality sign because the absolute value of a negative number is positive: \[ -d < 14 \] \[ d > -14 \] So the solution for the inequality \( 3|d| + 5 < 47 \) is: \[ -14 < d < 14 \] This means that \( d \) must be greater than -14 and less than 14.
Solving Absolute Value Inequalities
To solve the inequality \(|2x - 3| > |x + 3|\), you need to consider different cases based on the values of \(x\) which change the expressions inside the absolute value brackets. There are two main cases to consider for each absolute value expression: 1. When the expression inside the absolute value brackets is non-negative; 2. When the expression inside the absolute value brackets is negative. Let's deal with these step by step: For \(|2x - 3|\): - Case 1: \(2x - 3 \geq 0\), which simplifies to \(x \geq \frac{3}{2}\). - Case 2: \(2x - 3 < 0\), which simplifies to \(x < \frac{3}{2}\). For \(|x + 3|\): - Case 1: \(x + 3 \geq 0\), which simplifies to \(x \geq -3\). - Case 2: \(x + 3 < 0\), which simplifies to \(x < -3\). Combining these cases, you have to solve the inequality under the following scenarios: 1. \(x \geq \frac{3}{2}\) and \(x \geq -3\) (since \(x \geq \frac{3}{2}\) covers \(x \geq -3\)). This simplifies the inequality to \(2x - 3 > x + 3\): \(x > 6\). 2. \(x < \frac{3}{2}\) and \(x \geq -3\): Here we have to take two subcases because of the range of \(x\): a. \( -3 \leq x < \frac{3}{2} \), therefore the inequality becomes \(3 - 2x > x + 3\): \(-3x > 0\), which simplifies to \(x < 0\). b. \( x < -3 \), therefore the inequality becomes \(3 - 2x > 3 - x\): \(x < 0\), which always holds true for \(x < -3\). Putting it all together, there are two intervals for \(x\) that solve the inequality: - \(x < 0\) for the range between \(-3\) and \(\frac{3}{2}\), - \(x > 6\). The final solution to the inequality \(|2x - 3| > |x + 3|\) is: \[x < 0 \text{ or } x > 6.\]
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