Algebraic Solution of Linear Inequalities in One Variable

To isolate \(x\), subtract \(12\) from both sides: \(x+12-12\ge 25-12\), giving \(x\ge 13\). The inequality sign remains unchanged when the same number is subtracted from both sides. Adding \(12\) does not isolate \(x\). Exam tip: use the inverse operation of the term added to or subtracted from the variable.

To test whether a particular number satisfies an inequality, substitute that number for the variable and then evaluate both sides. The inequality here is strict: 2x-3<9. A strict less-than sign does not allow equality; the left side must be genuinely smaller than the right side.

Put x=6 into the expression: 2(6)-3=12-3=9. The statement then becomes 9<9, which is false because 9 equals 9 rather than being less than 9. Therefore option A is correct: the statement will be false. Solving the inequality also confirms this: 2x-3<9 gives 2x<12 and x<6. Since x=6 is the boundary value and is excluded, it cannot satisfy the original inequality.

The governing idea is substitution and verification of an inequality. Replace x by 6 in the left-hand side: 6x + 4 = 6(6) + 4 = 36 + 4 = 40. The inequality therefore becomes 40 ≤ 40. This statement is true because the symbol ≤ means “less than or equal to,” so equality at the boundary is permitted. Hence x = 6 satisfies the inequality, and option A is correct. Option B is wrong because 40 is not greater than itself; it is equal. Option C incorrectly gives a complete solution set when the question asks only for checking one value, while option D changes the symbol without any algebraic reason.

Add 9 to both sides: x − 9 + 9 > −4 + 9, so x > 5. Adding the same number to both sides does not change the direction of the inequality.

Subtracting (14) from both sides gives (x\le -16). Even with a negative answer, subtraction does not change the sign.

Subtracting 11 from both sides gives −2x < −6. Dividing both sides by −2 reverses the inequality sign, so x > 3. Option B incorrectly keeps the sign unchanged. Exam tip: Always reverse the inequality sign when multiplying or dividing by a negative number.

The direct answer is option A: x≤3. Solve 6-5x≥-9 carefully. Subtract 6 from both sides: -5x≥-15. Now divide by -5. Division by a negative number reverses the inequality sign, so x≤3. A is correct and includes x=3; substitution gives 6-15=-9, and -9≥-9 is true. B, x≥3, forgets the reversal and gives the wrong direction. C, x<3, incorrectly excludes x=3 even though equality works. D, x>3, has both the wrong direction and the wrong endpoint. The reversal is essential: multiplying or dividing an inequality by a negative number changes which side is larger. A number-line view also helps: all values at or left of 3 work. Memory cue: subtracting 6 is harmless, but dividing by -5 flips ≥ into ≤.

In option A, subtracting \(2x+3\) from both sides gives \(0<0\), which is never true. Hence, it has no real solution. In option D, \(0\geq0\) is true, so every real number satisfies it. Exam tip: cancel identical terms and check the resulting statement.

The governing concept is solving a linear inequality while preserving its direction. First distribute 2: 2x + 14 ≤ 4x + 6. Subtract 2x from both sides to obtain 14 ≤ 2x + 6, and then subtract 6 to get 8 ≤ 2x. Dividing by the positive number 2 does not reverse the inequality, so 4 ≤ x, which is conventionally written as x ≥ 4. Therefore option A is correct. Option B reverses the solution direction, while options C and D incorrectly replace the inclusive symbol with a strict one. The endpoint x = 4 must be included because substitution gives equality: 2(11) = 22 and 4(4) + 6 = 22.

Expand the brackets first: \\(9x-9>6x+12\\). Subtracting \\(6x\\) from both sides gives \\(3x-9>12\\), and adding 9 gives \\(3x>21\\). Dividing by 3 yields \\(x>7\\), so option A is correct. Option B reverses the inequality incorrectly, while options C and D include 7, which is not allowed in the strict inequality \\(>\\). Exam tip: the inequality sign changes direction only when dividing or multiplying by a negative number.

The statement translates to \(x+18>50\). Subtracting 18 from both sides gives \(x>32\), so option A is correct. ‘Greater than 50’ means strictly more than 50, so \(x\ge 32\) is not correct because it includes \(x=32\). In exams, use \(>\) for ‘greater than’ and \(\ge\) for ‘greater than or equal to.’

This word problem uses the boundary principle for a linear inequality. Starting with x + 12 ≥ 45, subtract 12 from both sides: x ≥ 45 − 12 = 33. The symbol ≥ includes equality, so x may be exactly 33 as well as any larger number. Therefore the minimum possible value is 33, making option A correct. The value 32 is too small because 32 + 12 = 44, which does not reach 45. The value 45 is allowed but is not the least possible value, and 57 is also larger than necessary. No reversal of the inequality occurs because only subtraction is performed, and subtracting the same number from both sides preserves the order.

‘Three times’ means \(3x\), so the inequality is \(3x<27\). Dividing both sides by the positive number 3 gives \(x<9\). Option C is incorrect because ‘less than’ does not include 9. Exam tip: the inequality sign reverses only when dividing or multiplying by a negative number.

‘At least’ means \(\ge\), so the inequality is \(x-7\ge 20\). Adding 7 to both sides gives \(x\ge 27\). Option B is incorrect because it uses the strict sign \(>\); in fact, \(x=27\) is also a valid solution. Exam tip: remember that ‘at least’ corresponds to \(\ge\), while ‘at most’ corresponds to \(\le\).

The governing concept is undoing operations in reverse order. In 2x + 3 ≤ 15, the addition of 3 is the outer operation applied after multiplication by 2. Therefore first subtract 3 from both sides: 2x + 3 − 3 ≤ 15 − 3, giving 2x ≤ 12. Next divide both sides by the positive number 2, obtaining x ≤ 6. Thus option A gives the correct order and the correct solution. Option B moves farther from isolating x by adding 3. Option C divides before removing the added constant and does not correctly undo the expression as written. Option D applies unrelated operations. Because the divisor is positive, the inequality sign remains ≤.

The key rule is that multiplying or dividing an inequality by a negative number reverses its direction. Begin with −x + 9 > 1. Subtract 9 from both sides to get −x > −8. To change −x into x, divide both sides by −1. Since −1 is negative, the sign reverses: x < 8. Therefore option A is correct. A quick check confirms the result: if x = 7, then −7 + 9 = 2 > 1; if x = 8, the expression equals 1, which is not strictly greater than 1. Thus 8 is excluded. Options B and D fail to reverse the sign, while option C incorrectly includes the equality case.

The governing concept is isolating the variable while preserving the inequality direction under positive operations. Start with 13 < 3x + 1. Subtract 1 from both sides to obtain 12 < 3x. Divide both sides by the positive number 3; the sign does not reverse, so 4 < x, equivalently x > 4. Hence option A is correct. The inequality is strict, so x = 4 is not included: substituting it gives 13 < 13, which is false. Option B reverses the direction, while options C and D incorrectly introduce an equality sign. A test value such as x = 5 gives 13 < 16, confirming that values greater than 4 belong to the solution set.

The governing idea is to isolate x and then rewrite the result in standard order. From 16 ≥ 2x + 4, subtract 4 from both sides: 12 ≥ 2x. Divide by the positive number 2, so the direction remains unchanged: 6 ≥ x. Reversing the order of the two sides gives x ≤ 6. Therefore option A is correct. The equality is included because the original symbol is ≥; indeed, x = 6 produces 16 ≥ 16, which is true. Option B reverses the solution region, while C and D incorrectly exclude or reverse the boundary. The phrase “standard form” simply means writing the variable on the left, not changing the mathematical meaning.

Subtracting \(2x\) from both sides gives \(5x-12<18\). Adding 12 to both sides gives \(5x<30\), and dividing by 5 gives \(x<6\). Since 5 is positive, the inequality sign does not reverse. In exams, remember to reverse the sign only when multiplying or dividing by a negative number.

Subtracting (5) gives (-3x\le -15), so division by (-3) gives (x\ge 5). Dividing by a negative number reverses the sign.

The governing concept is clearing a positive denominator without changing the inequality direction. Begin with (x − 3)/5 > 4. Since 5 is positive, multiply both sides by 5 safely: x − 3 > 20. Add 3 to both sides to obtain x > 23. Thus option A is correct. The inequality remains strict because the original symbol is >, so x = 23 is not included; substituting it gives 20/5 = 4, not a value greater than 4. Option B reverses the direction without justification, while C and D incorrectly include equality or give the opposite region. If x = 24, the left side is 21/5 = 4.2, confirming that values above 23 satisfy the inequality.

The inequality gives (4x\le 16), so (x\le 4) and the greatest integer is (4). The sign (\le) includes the boundary value.

Subtracting (3) from both sides of (x+3>7) gives (x>4). In exams remember that subtracting the same number does not change the sign.

Adding (5) to both sides of (x-5\le 2) gives (x\le 7). Addition keeps the inequality sign unchanged.

Dividing (3x\ge 18) by (3) gives (x\ge 6). The sign stays the same when dividing by a positive coefficient.