Algebraic Solution of Linear Inequalities in One Variable

Multiplying every part of the inequality by the positive number 2 does not reverse the inequality signs: \(-2<x-4\le12\). Adding 4 to every part gives \(2<x\le16\). Therefore, option A is correct. In option B, the inequality signs at both endpoints are incorrect. Exam tip: reverse inequality signs only when multiplying or dividing by a negative number.

From \(-10<4x+2\le18\), subtract 2 throughout to get \(-12<4x\le16\). Since 4 is positive, dividing by 4 does not reverse either inequality sign: \(-3<x\le4\). The greatest integer in this interval is \(4\). For \(x=5\), \(4x+2=22\), which is greater than 18. Exam tip: inequality signs reverse only when multiplying or dividing by a negative number.

Simplifying the left-hand side gives \(3(2x-1)-2(x+5)=6x-3-2x-10=4x-13\). Thus, \(4x-13\le x+4\), so \(3x\le17\). Since 3 is positive, dividing both sides by 3 does not reverse the inequality sign, giving \(x\le\frac{17}{3}\). Therefore, option C is correct. Option D has the inequality sign reversed. Exam tip: Reverse an inequality only when multiplying or dividing by a negative number.

Multiplying every term by (\text{LCM}=6) gives (x-13>1), so (x>14). In exams, remove fractions first and track the inequality sign carefully.

Solving both parts separately gives (x\ge \frac{3}{5}) and (x>-\frac{2}{3}), so the common solution is (x\ge \frac{3}{5}). For compound inequalities, always take the intersection.

Simplifying gives (7-6x<10-5x), so (7<10+x) and hence (x>-3). Since the exact interval is not listed, this item tests careful option checking.

Multiplying by (\text{LCM}=10) gives (2x-4\le 1-2x), so (4x\le 5) and (x\le \frac{5}{4}). Do not choose an approximate option when the exact answer is absent.

The left side is (x-14), so (x-14\ge 5-\frac{x}{2}) gives (\frac{3x}{2}\ge 19) and (x\ge \frac{38}{3}). Sign errors while opening brackets are very common in exams.

Simplifying gives (1-2x>4x-11), so (12>6x) and (x<2). While writing intervals, pay attention to open and closed endpoints.

Multiplying by (\text{LCM}=4) gives (21-3x\le 8x+14), so (x\ge \frac{7}{11}). After removing fractions, move terms to the correct side carefully.

Multiplying every term by (12) gives (5x-31\leq 0). In exams keep the inequality sign unchanged when multiplying by a positive number.

Simplification gives (-5x>-3) and division by a negative reverses the sign. This is the most common exam error.

Given \( -3\leq 2x+5<11 \), subtract 5 from all three parts to get \( -8\leq 2x<6 \). Dividing all parts by the positive number 2 does not reverse the inequality signs, giving \( -4\leq x<3 \). Hence, option A is correct. In option C, the open and closed endpoints have been reversed. Exam tip: Inequality signs remain unchanged when dividing by a positive number, but reverse when dividing by a negative number.

Multiplying by (12) gives (6x-9\geq 2x+10) and (x\geq \frac{19}{4}). Therefore the smallest integer is (7).

Expanding the brackets gives \(10-5x\leq 12-3x-8\), i.e. \(10-5x\leq 4-3x\). Adding \(5x\) to both sides gives \(10\leq 4+2x\), so \(6\leq 2x\) and hence \(x\geq 3\). The closest distractor is \(x>3\), but it excludes \(x=3\); the equality case is included because the original sign is \(\leq\). Exam tip: dividing an inequality by a positive number does not reverse its sign.

Each linear inequality in one variable gives a ray, all real numbers, or an empty set. In a system, all inequalities must hold simultaneously, so we take the intersection of their solution sets. An intersection of rays or intervals is again an interval, a ray, a single point, all real numbers, or the empty set. Option B is incorrect because an AND-system cannot produce two separated non-empty intervals. Exam tip: use intersection for “and” and union for “or”.

Expanding the brackets gives the left-hand side as \(2x-2+3x+6=5x+4\). Thus, \(5x+4\geq 4x+11\) gives \(x+4\geq 11\) after subtracting \(4x\) from both sides, so \(x\geq 7\). The option \(x>7\) is not correct because at \(x=7\), both sides are equal, and \(\geq\) includes equality. Exam tip: When an inequality includes equality, always test the boundary value.

Multiply both sides by 10 to remove the decimals: \(4x-12\leq x+21\). Collecting the \(x\)-terms on one side gives \(3x\leq 33\), so \(x\leq 11\). The option \(x<11\) is not correct because \(x=11\) makes both sides equal, and equality is included in \(\leq\). Exam tip: multiplying or dividing an inequality by a positive number does not reverse its sign.

The LCM of the denominators 3, 4, and 6 is 12. Multiplying both sides of the inequality by 12 gives \(8x+4-3x+6\geq 2x+10\). Thus, \(5x+10\geq 2x+10\), so \(3x\geq 0\), which gives \(x\geq 0\). The option \(x>0\) is incorrect because \(x=0\) also satisfies the inequality. Exam tip: the inequality sign does not change when multiplying by a positive number.

On simplifying, \(9-4x-4<2x-7\), so \(5-4x<2x-7\). Adding \(4x+7\) to both sides gives \(12<6x\). Since 6 is positive, dividing by 6 does not reverse the inequality sign, giving \(x>2\). The option \(x\geq2\) is incorrect because the original inequality is strict, so \(x=2\) is not included. Exam tip: An inequality sign remains unchanged when dividing by a positive number.

The first part gives (x\leq \frac{17}{7}) and the second gives (x>-4). Hence the combined solution is ((-4,\frac{17}{7}]).

Multiplying both sides by the positive number 4 does not reverse the inequality: \(-2(x-6)\leq 3x+2\). Thus, \(-2x+12\leq 3x+2\), so \(10\leq 5x\), giving \(x\geq 2\). The closest distractor is \(x>2\), but \(x=2\) makes both sides equal, so 2 must be included. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Solve the inequality: \(2x-5<11\Rightarrow 2x<16\Rightarrow x<8\). Taking natural numbers as \(1,2,3,\ldots\), the possible values of \(x\) are \(1,2,3,4,5,6,7\). Hence, there are 7 values. The answer would be 8 only under the convention that 0 is included in natural numbers; this question uses the common school convention \(\mathbb{N}=\{1,2,3,\ldots\}\). Exam tip: After solving an inequality, count values according to the specified number set.

Multiplying both sides by the positive number \(6\) does not change the inequality sign: \(36-2x\geq 6x-3\). Hence, \(39\geq 8x\), so \(x\leq \frac{39}{8}\). The option \(x<\frac{39}{8}\) is incorrect because \(x=\frac{39}{8}\) makes both sides equal, and \(\geq\) includes equality. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

From \(3x+a<12\), we get \(3x<12-a\), so \(x<\frac{12-a}{3}\). Since the given solution is \(x<2\), the boundary values must be equal: \(\frac{12-a}{3}=2\). Thus, \(12-a=6\), giving \(a=6\). For example, if \(a=5\), the solution would be \(x<\frac{7}{3}\), not \(x<2\). Exam tip: when a solution boundary is given, equate it to the boundary expression obtained after solving the inequality.