Algebraic Solution of Linear Inequalities in One Variable

First remove the bracket: \(4-(x-3)=4-x+3=7-x\). Thus, \(7-x\geq 2x+1\), which gives \(6\geq 3x\). Dividing both sides by the positive number \(3\) yields \(x\leq 2\), so option A is correct. Option B incorrectly reverses the inequality; the sign does not change when dividing by a positive number. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

\(\frac{x}{2}-\frac{x}{3}=\frac{3x-2x}{6}=\frac{x}{6}\), so \(\frac{x}{6}>1\). Multiplying both sides by the positive number 6 gives \(x>6\). Option C is incorrect because the original sign is \(>\), not \(\geq\); therefore, 6 itself is not included in the solution set. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number; here, 6 is positive.

Multiply both sides by 5 to clear the denominators. Since 5 is positive, the inequality sign does not change: \(3x+10\leq x+30\). Thus, \(2x\leq20\), giving \(x\leq10\). Therefore, option A is correct. Option B reverses the inequality incorrectly, while C and D wrongly exclude or change the endpoint \(x=10\). Exam tip: the inequality sign reverses only when multiplying or dividing by a negative number.

Multiplication or division by a negative number reverses the order on the number line, so the inequality sign must reverse. For example, multiplying 2 < 5 by −1 gives −2 > −5. Exam tip: flip the sign whenever you multiply or divide by a negative value.

The governing concept is solving a compound inequality by applying the same operation to all three parts. Starting with -6 ≤ 3x < 12, divide the entire chain by 3. Since 3 is positive, the directions of both inequality signs remain unchanged: -6/3 ≤ x < 12/3. Therefore -2 ≤ x < 4, which is option A. The left endpoint -2 is included because the left sign is ≤, while the right endpoint 4 is excluded because the right sign is <. Option B reverses the endpoint inclusion, option C excludes both endpoints, and option D incorrectly includes the upper endpoint. Testing the boundaries confirms the result: x = -2 works, but x = 4 does not.

Solving the inequality gives \(7x-4\geq 3x+20\), so \(4x\geq 24\) and hence \(x\geq 6\). Therefore, the least integer solution is 6. Option 5 is incorrect because substituting \(x=5\) gives \(31\geq 35\), which is false. Exam tip: the inequality sign does not reverse when multiplying or dividing by a positive number.

Expanding the bracket gives \(5-2(3x-1)=7-6x\), so \(7-6x\leq 15\) and hence \(-6x\leq 8\). Dividing by \(-6\) reverses the inequality sign, giving \(x\geq -\frac{4}{3}\). Option B is incorrect because it fails to reverse the sign. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

The least common multiple of the denominators 5 and 2 is 10, which is positive, so the inequality sign does not change. Multiplying by 10 gives \(2(4x-3)<5(x+6)\), or \(8x-6<5x+30\). Thus, \(3x<36\), and hence \(x<12\). Therefore, option A is correct. Exam tip: the inequality sign reverses when both sides are multiplied or divided by a negative number.

Subtracting 1 from all parts of \(3<2x+1\leq 13\) gives \(2<2x\leq 12\). Dividing by the positive number 2 yields \(1<x\leq 6\). Thus the integer values are \(x=2,3,4,5,6\), giving a total of 5 values. Remember that \(x=1\) is excluded, while \(x=6\) is included because the right-hand inequality uses \(\leq\).

Expanding the bracket gives \(9-4x-8>x-14\), or \(1-4x>x-14\). Simplifying further yields \(15>5x\), so \(x<3\). Therefore, option A is correct. Option C is incorrect because the original inequality is strict, so \(x=3\) is not included. Exam tip: the inequality sign reverses when both sides are divided or multiplied by a negative number, but not when divided by the positive number 5.

“Not less than −1” means greater than or equal to −1, so the inequality is 2x − 9 ≥ −1. Adding 9 gives 2x ≥ 8; dividing by positive 2 gives x ≥ 4. The equality case x = 4 is included.

Use the order-preserving operations for a compound inequality. Begin with -3 ≤ (x + 4)/2 < 5. Multiply all three parts by 2, a positive number, so the signs remain unchanged: -6 ≤ x + 4 < 10. Subtract 4 from all parts to obtain -10 ≤ x < 6. Therefore option A is correct. The lower endpoint is included because the original left sign is ≤, and the upper endpoint is excluded because the original right sign is <. Options B and D wrongly exclude -10, while option C incorrectly includes 6. Substitution at the endpoints verifies the result: x = -10 gives equality on the left, whereas x = 6 makes the right inequality strictness fail.

Simplifying the left side gives \(2x-10+3=2x-7\), and simplifying the right side gives \(3x-12-7=3x-19\). Thus, \(2x-7\geq 3x-19\). Subtracting \(2x\) from both sides gives \(-7\geq x-19\); adding 19 gives \(12\geq x\), or \(x\leq 12\). Therefore, option A is correct. Option B reverses the direction incorrectly. Exam tip: the inequality sign changes direction only when multiplying or dividing by a negative number.

Subtracting 13 from both sides gives \\(-3x\leq -9\\). Dividing by \\(-3\\) reverses the inequality sign, so \\(x\geq 3\\). Option A 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 both sides by a negative number.

Starting with \(6x-7>2x+17\), subtracting \(2x\) from both sides gives \(4x-7>17\). Adding 7 and then dividing by the positive number 4 gives \(4x>24\), hence \(x>6\). Therefore, option D is correct. Option C, \(x\geq6\), is incorrect because the original inequality is strict and does not include equality. Exam tip: dividing by a positive number does not reverse the inequality sign.

Multiplication or division by a negative number reverses order: if \(a<b\), then \(-a>-b\). Thus, multiplying \(x>-2\) by \(-1\) gives \(-x<2\). Exam tip: reverse the inequality sign whenever you multiply or divide by a negative value.

Expanding the brackets gives \(10x+5-3x<26\). Combining like terms, \(7x+5<26\), so \(7x<21\). Dividing by the positive number 7 gives \(x<3\), making option B correct. Option C, \(x≤3\), is incorrect because the original inequality is strict; \(x=3\) does not satisfy it. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number; since 7 is positive here, the sign remains unchanged.

Adding 1 to all three parts of \(2\le 3x-1<11\) gives \(3\le 3x<12\). Dividing throughout by 3 gives \(1\le x<4\). Thus, 1 is included and is written with a square bracket, whereas 4 is excluded and is written with a round bracket. Therefore, the interval is \([1,4)\). Exam tip: use a square bracket for an included endpoint marked by \(\le\) or \(\ge\), and a round bracket for an excluded endpoint marked by \(<\) or \(>\).

Adding 3 to both sides of \(4x-3<21\) gives \(4x<24\). Dividing by the positive number 4 yields \(x<6\). Therefore, the greatest integer less than 6 is 5. Option 6 is incorrect because a strict inequality does not include the boundary value \(x=6\). Exam tip: In inequalities with \(<\) or \(>\), the endpoint is not included.

Solving the inequality gives \(5x+1\leq 26\Rightarrow 5x\leq 25\Rightarrow x\leq 5\). Natural numbers begin with \(1\), so the possible values are \(1,2,3,4,5\), giving a total of 5 values. The answer 6 would result only if \(0\) were included as a natural number; here, the usual school convention takes natural numbers to begin with 1. Exam tip: After solving an inequality, list the allowed values and count them carefully.

The direct answer is option A: x<3. Begin with 18-5x>3. Subtract 18 from both sides: -5x>-15. Divide by -5. Because division by a negative number reverses the sign, x<3. A is correct. Test x=2: 18-10=8>3, so it works. The boundary x=3 is excluded because 18-15=3>3 is false. B, x>3, has the wrong direction; x=4 gives -2>3, false. C, x≤3, incorrectly includes x=3 even though strict inequality fails there. D, x≥3, has the wrong direction and includes invalid values. The most common error is dividing by -5 while leaving > unchanged. A number-line answer has an open circle at 3 and shading to the left. Memory cue: a strict > remains strict, and a negative division flips it to <.

Multiplying both sides by the positive number 4 does not reverse the inequality: \(3x-2>28\). Thus, \(3x>30\), giving \(x>10\). Option D is incorrect because the original inequality is strict; when \(x=10\), \(\frac{30-2}{4}=7\), and \(7>7\) is false. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Since 3 is positive, multiplying both sides by 3 gives \(7-x\leq 6\). Thus, \(-x\leq -1\). Multiplying both sides by \(-1\) reverses the inequality sign, giving \(x\geq 1\). Therefore, option B is correct. Exam tip: Always reverse the inequality sign when multiplying or dividing an inequality by a negative number.

Multiplying by (4) gives (2x+10+x-1\geq 24). So (3x\geq 15), giving (x\geq 5), so recheck arithmetic carefully.

Multiply both sides by 4 to clear the denominators. Since 4 is positive, the inequality sign does not change: \(2(x+5)+(x-1)\geq 24\). Thus, \(3x+9\geq 24\), so \(3x\geq 15\) and \(x\geq 5\). Therefore, option A is correct.