Algebraic Solution of Linear Inequalities in One Variable

This is a one-variable linear inequality, and the least common denominator is 6. Since 6 is positive, multiplying every term by 6 preserves the ≤ sign. We obtain 3(x−1)+2(x−2) ≤ x+4. Expanding gives 3x−3+2x−4 ≤ x+4, so 5x−7 ≤ x+4. Subtracting x and adding 7 produce 4x ≤ 11, and division by the positive number 4 gives x ≤ 11/4. Therefore option A is correct. Option B reverses the inequality, option C incorrectly excludes the endpoint even though equality is permitted, and option D comes from an arithmetic error. The endpoint is included because the original relation is ≤.

The governing rule is to clear a positive denominator before collecting the variable terms. Multiplying both sides by 3 preserves the inequality because 3>0: 2−5x ≥ 12x−21. Add 21 to both sides to obtain 23−5x ≥ 12x, then add 5x to get 23 ≥ 17x. Dividing by the positive coefficient 17 yields x ≤ 23/17. Hence option A is correct. Option B reverses the direction without a negative divisor, option C uses a strict sign although equality is allowed, and option D changes the sign of the boundary incorrectly. Checking x=0 satisfies the original inequality, and 0≤23/17, while a value larger than 23/17 fails, confirming the result.

The governing rule is that multiplying or dividing an inequality by a negative number reverses its inequality signs. Start with −4 ≤ 5 − 3x < 11 and subtract 5 throughout: −9 ≤ −3x < 6. Now divide every part by −3, reversing both signs: 3 ≥ x > −2. Writing the result in increasing order gives −2 < x ≤ 3. Thus option C is correct. The endpoint −2 is excluded because the original middle expression would equal 11, but the right inequality is strict. The endpoint 3 is included because the expression equals −4, and that boundary is non-strict. Options A and B retain only one-sided restrictions, while D incorrectly excludes x = 3.

For an ‘or’ statement, the combined solution is the union of the solution sets of the two inequalities. Solve the first inequality: 2x − 9 ≤ x − 4, so x ≤ 5. Solve the second: 5x + 3 > 18, hence 5x > 15 and x > 3. The union is (−∞, 5] ∪ (3, ∞). Every real number belongs to this union: numbers at most 5 satisfy the first condition, and numbers greater than 5 satisfy the second condition. Therefore the combined solution is all real numbers, so option D is correct. Option A gives only the first part, option B gives only the second part, and option C wrongly treats the conditions as contradictory.

Because 5 is positive, multiplying the entire compound inequality by 5 does not reverse either sign: −15 ≤ 7 − 2x < 5. Subtract 7 throughout to obtain −22 ≤ −2x < −2. Dividing by −2 reverses both inequalities, giving 11 ≥ x > 1. In standard increasing order, this is 1 < x ≤ 11, so option A is correct. The lower endpoint 1 is excluded because the original upper inequality is strict: at x = 1, the fraction equals 1. The upper endpoint 11 is included because the fraction equals −3 there, and that boundary is non-strict. Option B reverses the endpoint inclusion, while C and D retain only incorrect one-sided portions.