Algebraic Solution of Linear Inequalities in One Variable

Adding 2 to both sides of \(x-2<5\) gives \(x<7\). Since \(x\geq 2\) and \(x\) is an integer, the possible values are the integers from 2 through 6: \(\{2,3,4,5,6\}\). Option B is incorrect because 7 is not allowed by the strict inequality \(x<7\). Exam tip: a strict inequality, such as \(<\) or \(>\), excludes its boundary value.

Subtracting \(2x\) from both sides gives \(3x+1>16\). Subtracting 1 and then dividing by the positive number 3 gives \(x>5\). Since the original inequality is strict, \(x=5\) is not included; therefore, \(x\geq5\) is incorrect. Exam tip: Reverse the inequality sign only when multiplying or dividing by a negative number; here, 3 is positive.

In option A, subtracting 1 from both sides gives \(x+1\geq 0\Rightarrow x\geq -1\). Option C gives \(x\leq -1\), which has the opposite direction. Exam tip: adding or subtracting a positive number does not reverse the inequality sign.

In \\(7x\geq -21\\), dividing both sides by the positive number 7 does not reverse the inequality sign. Thus, \\(x\geq -3\\). Option B is incorrect because the sign would reverse only when multiplying or dividing by a negative number. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

Dividing both sides by \(-7\) reverses the inequality sign because the divisor is negative: \(x\geq \frac{28}{-7}=-4\). Hence, \(x\geq -4\) is correct. Option B is incorrect because it does not reverse the sign. Exam tip: Always reverse the inequality sign when multiplying or dividing an inequality by a negative number.

The governing rule is that solving a linear inequality follows the usual equation steps, except that multiplying or dividing by a negative number reverses the inequality sign. Start with 5−2x<1. Subtract 5 from both sides to obtain −2x<−4. Now divide both sides by −2; because the divisor is negative, the sign changes from < to >, giving x>2. A quick check supports this result: x=3 gives 5−6=−1, and −1<1 is true, whereas x=1 gives 3<1, which is false. Option B forgets the reversal, while C and D use the wrong boundary or direction. Thus A is correct.

The key concept is reversing an inequality when dividing by a negative number. Test option A directly: −2x≥2. Divide both sides by −2, and reverse the sign, obtaining x≤−1. Thus its solution set is exactly the one requested. Option B would give x≥−1 because the sign reverses during the same division. Option C gives x≥−1 because division is by positive 2 and the direction stays unchanged. Option D gives x<−1, which excludes the endpoint −1 and therefore is not equivalent to x≤−1. The endpoint is important: x=−1 satisfies option A as equality. Hence option A is the only correct inequality.

Solve the strict linear inequality step by step. Starting with 2x−7>5, add 7 to both sides to obtain 2x>12. Dividing by the positive number 2 preserves the direction, giving x>6. Thus x=6 is the boundary value, but it is excluded because the original sign is > rather than ≥. Substitution confirms this: 2(6)−7=5, and 5>5 is false. For x=7, 8, or 9, the left side is respectively 7, 9, or 11, all of which are greater than 5. Therefore only option A is not a solution. The strict sign and endpoint exclusion are the decisive points.

The governing concept is translating a verbal double condition into a compound inequality and then simplifying all three parts consistently. “2x+3 is greater than 1 and less than 11” becomes 1<2x+3<11. Subtract 3 from every part: −2<2x<8. Divide all parts by the positive number 2, so the inequality directions remain unchanged: −1<x<4. Both endpoints are excluded because the original statements say “greater than” and “less than,” not “at least” or “at most.” Option B incorrectly includes −1, option C incorrectly includes 4, and option D reverses the resulting range. Therefore option A is correct.

The governing ideas are solving a compound linear inequality and then restricting the result to integers. Begin with 1≤2x−1<9. Add 1 to every part to get 2≤2x<10. Divide by positive 2, preserving both inequality signs: 1≤x<5. Over the real numbers this is an interval, but x is specified to belong to Z, the integers. The integers satisfying 1≤x<5 are 1, 2, 3, and 4, so the solution set is {1,2,3,4}, option A. Zero is too small, 5 is excluded by the strict upper bound, and option D omits the valid endpoint value 1. Hence A is correct.

The governing concept is the sign reversal that occurs when dividing an inequality by a negative number. Start with 4−7x>18 and subtract 4 from both sides, obtaining −7x>14. Dividing by −7 reverses the inequality from > to <, so x<−2. A boundary check shows that x=−2 gives 4−7(−2)=18, but equality is not allowed; therefore −2 must be excluded. A value such as x=−3 gives 25>18 and confirms the direction. Option B forgets the reversal, while C and D use a positive boundary or incorrect direction. Thus the exact solution is x<−2, option A.

The governing rule is that an inequality reverses direction when both sides are divided or multiplied by a negative number. Start with -5x + 4 ≤ -16. Subtract 4 from both sides to obtain -5x ≤ -20. Now divide by -5. Because -5 is negative, the symbol changes from ≤ to ≥, giving x ≥ 4. Hence option B is correct. A quick check confirms this: x = 4 gives -5(4) + 4 = -16, so the endpoint is included; x = 5 gives -21, which is also less than -16. Options A and C retain or assign an incorrect direction, while option D has the wrong boundary value. The solution set is [4, infinity).

A compound inequality must be transformed on all three parts at the same time. Begin with -3 ≤ 2x + 1 < 7. Subtract 1 throughout to get -4 ≤ 2x < 6. Divide every part by the positive number 2; the inequality signs remain unchanged, giving -2 ≤ x < 3. The real-number solution is the interval from -2 inclusive to 3 exclusive. Restricting this interval to integers gives -2, -1, 0, 1, and 2, so option A is correct. The value 3 is excluded because the right inequality is strict, and -2 is included because the left inequality uses ≤. Option B omits -2 and includes 3, option C incorrectly includes 3, and option D incorrectly includes -3.

The essential concept is sign reversal when an inequality is multiplied by a negative quantity. Starting with (3 - x)/(-4) < 2, multiply both sides by -4. Since -4 is negative, the less-than sign becomes greater-than: 3 - x > -8. Subtract 3 from both sides to get -x > -11. Multiplying by -1 again reverses the sign, so x < 11. Thus option A is correct. The endpoint 11 is not included because the original inequality is strict; substituting x = 11 gives 2 < 2, which is false. A value such as x = 0 gives -3/4 < 2, confirming the direction. Options B and D reverse the final direction, while C uses the incorrect boundary -5.

To solve the inequality, subtract 2x from both sides. This gives 5≤9, which is a true numerical statement independent of x. Therefore every real value of x satisfies the original inequality. This type of result is called an identity in the context of the inequality: the variable terms cancel and the remaining statement is always true. For example, x=0 gives 5≤9, x=10 gives 25≤29, and x=−7 gives −9≤−5; all work. Hence option A is correct. Option B would be correct only if cancellation produced a false statement such as 5≤3. Options C and D incorrectly impose restrictions on x that do not arise from the simplified inequality.

This is a linear inequality in which the variable terms cancel. Subtract x from both sides: x + 3 < x − 2 becomes 3 < −2. That remaining statement is false, and its truth does not depend on the value of x. Therefore the original inequality has no real solution, so option D is correct. Options A, B, and C all claim that one or more values of x work, but substituting any real number produces the same impossible comparison 3 < −2. It is important not to treat cancellation of x as proof that every value works; the constant statement left after cancellation must first be checked.

Apply equivalent operations carefully, especially division by a negative number. From 5 − 2x > −11, subtract 5 from both sides to obtain −2x > −16. Dividing by −2 reverses the inequality sign because multiplication or division by a negative number reverses order. Thus x < 8, making option A correct. A quick check confirms this: if x = 0, the original expression gives 5 > −11, which is true, and 0 is indeed less than 8. Option B results from forgetting the reversal, option C changes the magnitude incorrectly, and option D includes the endpoint 8, although x = 8 gives equality rather than a strict greater-than result.

The governing method is to perform equivalent operations on both sides of a linear inequality. Expanding the brackets gives 3x − 6 ≤ 2x + 10. Subtract 2x from both sides: x − 6 ≤ 10. Adding 6 to both sides then gives x ≤ 16. No sign reversal occurs because only addition and subtraction were used; equivalently, any division involved would be by a positive coefficient. The equality value is included: substituting x = 16 gives 3(14) = 42 and 2(21) = 42, so the original non-strict inequality is satisfied. Option B reverses the direction, option C removes the valid endpoint, and option D has both errors. Therefore option A is correct.

The governing principle is that ordinary algebraic operations preserve an inequality, while multiplication or division by a negative number reverses its sign. Expand the left side: 5(2x − 1) = 10x − 5, so 10x − 5 < 3x + 16. Subtract 3x from both sides to obtain 7x − 5 < 16. Add 5, giving 7x < 21. Since 7 is positive, division by 7 does not reverse the sign, and the result is x < 3. The boundary is excluded because the original symbol is strictly less than; x = 3 makes both sides equal to 25. Thus option A is correct, while B reverses the direction and C and D incorrectly include 3.

Adding 5 to both sides gives \(3x<15\). Dividing by 3 gives \(x<5\); since 3 is positive, the inequality sign remains unchanged. Option C and option D incorrectly include equality, while option B reverses the inequality. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Subtracting 7 from both sides gives \\(-2x\geq -6\\). Dividing by \\(-2\\) reverses the inequality sign, so \\(x\leq 3\\). Option B is incorrect because it does not reverse the sign after division by a negative number. Exam tip: Always reverse an inequality sign when multiplying or dividing by a negative number.

Subtracting 2 from both sides gives \(\frac{x}{4}\leq 3\). Multiplying both sides by 4 yields \(x\leq 12\). Since 4 is positive, the inequality sign remains unchanged, so option A is correct. Exam tip: the inequality sign reverses only when both sides are multiplied or divided by a negative number.

Solving the inequality \(5x+8>3x+20\), subtract \(3x\) from both sides to obtain \(2x>12\). Dividing by the positive number 2 gives \(x>6\), so option A is correct. Option C is incorrect because it includes \(x=6\), but at \(x=6\) both sides are equal, whereas the inequality is strict. Exam tip: reverse the inequality sign whenever you multiply or divide by a negative number.

Solve stepwise: \(2(x-3)\geq x+4\). Expand to get \(2x-6\geq x+4\). Subtract \(x\) from both sides: \(x-6\geq 4\). Add 6 to both sides: \(x\geq 10\). Thus the solution is \(x\geq 10\). Option C (\(x>10\)) is incorrect because the original inequality allows equality. Options B and D give the wrong direction. Exam tip: always expand and then isolate the variable; remember to reverse the inequality only when multiplying or dividing by a negative number.

Expanding the brackets gives \\(3x+6<2x+14\\). Subtracting \\(2x\\) from both sides yields \\(x+6<14\\), so \\(x<8\\). Therefore, option A is correct. Option C is incorrect because the original inequality is strict; when \\(x=8\\), both sides are equal, so the inequality is not satisfied. Exam tip: a strict inequality sign \\(<\\) excludes the boundary value.