Algebraic Solution of Linear Inequalities in One Variable

The solutions of a linear inequality in two variables, such as \(ax+by<c\), lie on one side of the line \(ax+by=c\). Hence, its graph represents a half-plane. The boundary line is included for \(\leq\) or \(\geq\), but excluded for \(<\) or \(>\). A straight line alone represents a linear equation, not a linear inequality. Exam tip: Use a test point such as \((0,0)\) to identify the correct side of the boundary line.

For \(x=4\), both \(4>-2\) and \(4\le 4\) are true; equality is allowed in \(\le\). The value \(x=-2\) fails the first inequality because it requires \(x>-2\), not \(x\ge -2\). Exam tip: test each option against both conditions separately.

Subtracting 4 from both sides of option A gives \(x<6\). By contrast, options B and D simplify to \(x>6\), and dividing option C by \(-1\) reverses its inequality sign, also giving \(x>6\).

Subtracting \(2x\) from both sides gives \(2x-3<9\). Adding \(3\) to both sides gives \(2x<12\), and dividing by the positive number \(2\) gives \(x<6\). Therefore, option A is correct. Option B is wrong because no division by a negative number occurs, so the inequality sign does not reverse. Exam tip: Reverse the inequality sign only when multiplying or dividing by a negative number.

Multiplying both sides by 2 gives \(5-x>6\), so \(-x>1\). Multiplying both sides by \(-1\) reverses the inequality sign, yielding \(x<-1\). Therefore, option A is correct. Exam tip: Always reverse the inequality sign when multiplying or dividing by a negative number.

Adding the same number, 2, to both sides of the given inequality \(x\ge 1\) gives \(x+2\ge 3\). Hence, option A is true for every value of \(x\ge 1\). Options B, C, and D are not always true; for example, when \(x=1\), B and D are false. Exam tip: Adding or subtracting the same number on both sides of an inequality does not reverse its sign.

Dividing both sides by 2 gives \(x-3>4\). Adding 3 to both sides yields \(x>7\), so option B is correct. In option C, the inequality sign has been incorrectly reversed; the sign reverses only when multiplying or dividing by a negative number. Exam tip: division by a positive number does not change the inequality sign.

Expanding the bracket gives \(3x+6\le 2x+11\). Subtracting \(2x\) from both sides yields \(x+6\le 11\), so \(x\le 5\). Option B reverses the inequality incorrectly; the sign does not change because we have not multiplied or divided by a negative number. Exam tip: expand brackets first and collect variable terms on one side.

For the inequality \(x-7\ge -2\), add 7 to both sides: \(x\ge -2+7=5\). Therefore, the solution is \(x\ge 5\). Option B is incorrect because adding a number does not reverse the inequality sign. Exam tip: reverse the inequality sign only when multiplying or dividing both sides by a negative number.

The governing concept is solving a one-variable linear inequality while reversing the sign when dividing by a negative number. Begin with 10 − 2x < 4. Subtract 10 from both sides to obtain −2x < −6. Now divide both sides by −2. Because −2 is negative, the inequality direction reverses: x > 3. The endpoint 3 is excluded, as can also be checked directly: at x = 3, the original left side is 10 − 6 = 4, and 4 < 4 is false. A value such as x = 4 gives 2 < 4, so it works. Hence option B is correct; the other choices either retain the wrong sign or calculate the boundary incorrectly.

The governing concept is counting integers in a compound inequality while treating closed and open endpoints correctly. The condition 2 ≤ x includes 2, whereas x < 7 excludes 7. Since x is restricted to the integers, list the values from the included lower endpoint up to, but not including, the upper endpoint: 2, 3, 4, 5, and 6. There are five values. Equivalently, the count is 7 − 2 = 5 because the lower endpoint is included and the upper endpoint is excluded. Option B is correct. Four would omit one valid integer, six would incorrectly include 7, and seven confuses the interval length with the number of integer points.

The governing concept is solving a linear inequality with the variable on both sides. Start with 5x + 4 > 2x − 8. Subtract 2x from both sides: 3x + 4 > −8. Then subtract 4 from both sides: 3x > −12. Dividing by the positive number 3 does not reverse the inequality, so x > −4. The boundary −4 is excluded because the original sign is strict; substituting x = −4 gives equality, not a value greater than the right side. A test value such as x = 0 confirms the result because 4 > −8 is true. Thus option A is correct; the other options reverse the direction or use the wrong boundary.

Adding 7 to both sides of \(2x-7\leq 9\) gives \(2x\leq 16\). Dividing by 2 yields \(x\leq 8\), so option A is correct. Option C is incorrect because the boundary value \(x=8\) is included. Exam tip: the inequality sign changes only when multiplying or dividing by a negative number.

Divide both sides by -4. Because -4 is negative, reverse the inequality sign: x > 20/(-4) = -5. Therefore, the solution is x > -5.

Adding 4 to both sides gives \(x\geq 10\). Thus, 10 and every real number greater than 10 are solutions, so the solution set is \([10,\infty)\). Option B excludes 10, but \(\geq\) includes the boundary value. Exam tip: use a square bracket at an included endpoint for \(\geq\) or \(\leq\).

Subtracting \(3x\) from both sides gives \(4x+3\geq 15\). Subtracting 3 and then dividing by the positive number 4 gives \(x\geq 3\). The inequality sign therefore remains unchanged. Option B incorrectly reverses the sign. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Subtracting 2 from both sides gives \(\frac{x}{3}<3\). Multiplying both sides by the positive number 3 yields \(x<9\), so option A is correct. The inequality sign remains unchanged when multiplying or dividing by a positive number. Exam tip: reverse the inequality sign only when multiplying or dividing by a negative number.

Subtracting 5 from both sides gives −2x > 6. Dividing by −2 reverses the inequality sign, so x < −3.

This is a compound linear inequality, so the same operation must be applied to all three parts. Begin with −3 ≤ x + 2 < 8. Subtract 2 from the left boundary, the middle expression, and the right boundary: −3 − 2 ≤ x + 2 − 2 < 8 − 2. This simplifies to −5 ≤ x < 6. Because only 2 was subtracted, the inequality signs retain their directions, and the inclusion of the left endpoint also remains unchanged. Thus option A is correct. Option B results from subtracting incorrectly, option C reverses endpoint inclusion without multiplying or dividing by a negative number, and option D shifts both boundaries incorrectly.

Starting with \(4x-5<2x+7\), subtract \(2x\) from both sides to get \(2x-5<7\). Adding 5 gives \(2x<12\), and dividing by the positive number 2 gives \(x<6\). Since the original sign is '<', the endpoint 6 is not included, so \(x\leq 6\) is incorrect. Exam tip: the inequality sign reverses only when multiplying or dividing by a negative number.

Solving the inequality, \(6x+1\leq 2x+17\) gives \(4x\leq 16\), so \(x\leq 4\). Thus, all integer solutions are 4 or less, and the greatest integer solution is 4. Option 5 is incorrect because substituting it gives \(31\leq 27\), which is false. In exams, remember to reverse the inequality sign when multiplying or dividing by a negative number.

Dividing both sides by the positive number \(3\) gives \(x-2>4\). Adding \(2\) to both sides yields \(x>6\), so option A is correct. Option C results from forgetting to add \(2\) after division by \(3\). Exam tip: Dividing or multiplying an inequality by a positive number does not reverse its sign.

Solve \(2(3x+1) \ge 4x+10\).
Expand: \(2(3x+1)=6x+2\).
So \(6x+2 \ge 4x+10\).
Subtract \(4x\): \(2x+2 \ge 10\).
Subtract \(2\): \(2x \ge 8\).
Divide by \(2\): \(x \ge 4\).
Hence the solution set is \(x \ge 4\).

The signs ≤ and ≥ include equality, so the boundary value is a solution. The signs < and > exclude it. Exam tip: use a filled dot on the number line for ≤ or ≥.

From \\(9-3x<0\\), we get \\(-3x<-9\\). Dividing both sides by \\(-3\\) reverses the inequality sign, giving \\(x>3\\). Option B results from forgetting this sign reversal, while the options with equality incorrectly include \\(x=3\\), which does not satisfy the original strict inequality. Exam tip: Reverse the inequality sign whenever you multiply or divide by a negative number.