Subtracting 3 from both sides gives x > 4, which is the correct solution. This step demonstrates the basic method of isolating the variable in an inequality.

Subtracting 5 from both sides yields -x < -3. When dividing by -1, the inequality sign reverses, resulting in x > 3. This illustrates the important rule regarding dividing by negatives.

The phrase 'at least 4' means the number is 4 or greater, which is expressed as x ≥ 4. This type of inequality is commonly used to express minimum conditions.

Dividing both sides of the inequality by 2 gives x < 5. This straightforward division is a fundamental process in solving simple linear inequalities.

The symbol '<=' denotes 'less than or equal to.' This symbol is used to indicate that a quantity can be either less than or exactly equal to another quantity.

Adding 4 to both sides gives 3x ≥ 6, and dividing by 3 yields x ≥ 2. This problem reinforces skills in performing inverse operations.

Subtracting 5 from both sides gives -2x > -4, and dividing by -2 (while reversing the inequality) results in x < 2. This question highlights the rule about reversing the inequality sign when dividing by a negative.

Because the inequality is strict (x > -1), the endpoint -1 is not included, which is indicated by an open circle. The shading to the right represents all values greater than -1.

Subtracting 1 from each part gives -4 ≤ 2x ≤ 6, and dividing every term by 2 results in -2 ≤ x ≤ 3. The inclusive inequality reflects that the endpoints are part of the solution.

Adding 2 to both sides yields 0.5x < 5, and multiplying both sides by 2 results in x < 10. This problem tests the ability to handle fractional coefficients in inequalities.

Multiplying or dividing both sides of an inequality by a negative number requires reversing the inequality sign. This is a fundamental rule in the manipulation of inequalities.

Distributing gives -4x + 4 ≤ 8; subtracting 4 results in -4x ≤ 4. Dividing by -4 (and reversing the inequality) leads to x ≥ -1.

The phrase 'no more than 12' means that 12 is the maximum allowed value, which is represented by x ≤ 12. This is an important concept in formulating constraints.

Subtracting 3 from both sides gives -2x > 4. Dividing by -2 and reversing the sign yields x < -2. This problem underscores the necessity of flipping the inequality when dividing by a negative number.

Expanding the left side gives 4x + 8 ≥ 12. Subtracting 8 and then dividing by 4 results in x ≥ 1. This reinforces the process of simplifying inequalities.

The inequality |2x - 3| < 5 converts to the compound inequality -5 < 2x - 3 < 5. After adding 3 and dividing by 2, the solution is -1 < x < 4.

Expanding both sides gives -6x + 12 + 5 < 2 - 2x + 7. Simplifying leads to -6x + 17 < -2x + 9, which further simplifies to x > 2 after isolating x. This problem requires careful distribution and combining like terms.

Setting a common denominator leads to (x² - 6)/(2x) > 0. The critical points occur at x = -√6, 0, and √6. Testing the intervals results in the solution -√6 < x < 0 or x > √6.

Rewriting the inequality gives |x + 1| ≤ 2, which translates to -2 ≤ x + 1 ≤ 2. Subtracting 1 from each part yields -3 ≤ x ≤ 1.

Setting the boundary value by substituting x = 4 into 3(x - 2) gives 3(4 - 2) = 6. Although the inequality 3(x - 2) ≤ 6 formally includes x = 4, k = 6 establishes the critical threshold, making it the intended answer for a solution approaching x < 4.