Example Question - inequality cases
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Solving Absolute Value Inequalities
The equation given in the image is an inequality involving an absolute value: \[ \left| \frac{x + 3}{2} \right| \geq 4 \] To solve this inequality, we must consider two cases because the absolute value expression can be either positive or negative: **Case 1: The expression inside the absolute value is positive or zero** \[ \frac{x + 3}{2} \geq 4 \] To solve this, multiply both sides by 2 to get rid of the denominator: \[ x + 3 \geq 8 \] Now, subtract 3 from both sides: \[ x \geq 5 \] **Case 2: The expression inside the absolute value is negative** \[ -\left( \frac{x + 3}{2} \right) \geq 4 \] Again, multiply both sides by 2: \[ -(x + 3) \geq 8 \] Distribute the negative sign: \[ -x - 3 \geq 8 \] Now, add 3 to both sides: \[ -x \geq 11 \] Finally, multiply both sides by -1 and remember to flip the inequality sign when multiplying or dividing by a negative number: \[ x \leq -11 \] Combining both cases gives us the solution set: \[ x \leq -11 \quad \text{or} \quad x \geq 5 \] These are the values of x that satisfy the original inequality.
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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