Example Question - compound inequality
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Solving a Compound Inequality
<p>Given the compound inequality \( \frac{3x+1}{2} \leq 5 + 2x \) and \( 1 + 5x \geq 8x \).</p> <p>First, solve \( \frac{3x+1}{2} \leq 5 + 2x \):</p> <p>Multiply both sides by 2 to eliminate the fraction: \( 3x+1 \leq 10 + 4x \).</p> <p>Subtract \( 3x \) from both sides: \( 1 \leq 10 + x \).</p> <p>Subtract 10 from both sides: \( -9 \leq x \) or \( x \geq -9 \).</p> <p>Now, solve \( 1 + 5x \geq 8x \):</p> <p>Subtract \( 5x \) from both sides: \( 1 \geq 3x \).</p> <p>Divide both sides by 3: \( \frac{1}{3} \geq x \) or \( x \leq \frac{1}{3} \).</p> <p>Combine the solutions: \( x \geq -9 \) and \( x \leq \frac{1}{3} \).</p> <p>So the solution to the compound inequality is \( -9 \leq x \leq \frac{1}{3} \).</p> <p>Thus, the correct answer is D. \( -9 \leq x \leq \frac{1}{3} \).</p>
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 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 \( |c| - 1 \geq 11 \), 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 \( |8r| + 2 > 42 \), 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 \( |8r| + 2 > 42 \).
Solving Absolute Value Inequality
To solve the inequality \( 3 \left| \frac{x+3}{2} \right| \leq 9 \), we can proceed as follows: 1. Divide both sides of the inequality by 3 to isolate the absolute value expression: \[ \left| \frac{x+3}{2} \right| \leq \frac{9}{3} \] \[ \left| \frac{x+3}{2} \right| \leq 3 \] 2. The absolute value inequality \( |A| \leq B \) where \( B \) is non-negative (B ≥ 0) can be split into two inequalities: \[ -B \leq A \leq B \] Applying this concept, we get: \[ -3 \leq \frac{x+3}{2} \leq 3 \] 3. Multiply all parts of the compound inequality by 2 to get rid of the fraction: \[ -6 \leq x+3 \leq 6 \] 4. Now subtract 3 from all parts of the compound inequality to solve for \( x \): \[ -6 - 3 \leq x+3 - 3 \leq 6 - 3 \] \[ -9 \leq x \leq 3 \] So the solution set for the inequality is \( x \in [-9, 3] \).
Solving a Compound Inequality with 'Or'
The image shows a compound inequality that consists of two parts connected by "or". The first part of the inequality is: \[ 6 + 7m \leq 6m - 5 \] The second part of the inequality is: \[ 3m - 7 < 5 + 6m \] To solve the first part of the inequality, we will isolate the variable \( m \) on one side: \[ 6 + 7m \leq 6m - 5 \] Subtract \( 6m \) from both sides: \[ 7m - 6m \leq -5 - 6 \] Combine like terms: \[ m \leq -11 \] Now to solve the second part of the inequality: \[ 3m - 7 < 5 + 6m \] Subtract \( 3m \) from both sides: \[ -7 < 5 + 3m \] Subtract \( 5 \) from both sides to isolate the variable term: \[ -12 < 3m \] Divide both sides by \( 3 \) to solve for \( m \): \[ -4 < m \] Now we have two parts to the solution, and since they are connected by "or," we will take the union of the two solutions. The solutions to the compound inequality are: \[ m \leq -11 \quad \text{or} \quad m > -4 \] This means that any \( m \) that is less than or equal to -11 or greater than -4 satisfies the original compound inequality.
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