Algebraic Solution of Linear Inequalities in One Variable

Correct answer: A. It is always convex.

Each linear inequality represents a half-plane, and every half-plane is convex. Their intersection therefore remains convex. It need not be bounded or contain the origin. Exam tip: use a test point to identify the valid side of each boundary.

Correct answer: A. सभी वास्तविक \(x\)

Subtracting \(x\) from both sides gives \(0<1\), which is always true. Therefore, \(x<x+1\) holds for every real value of \(x\). Options B and C impose unnecessary restrictions to positive or negative values, while option D is false because real solutions do exist. Exam tip: when identical variable terms cancel and leave an always-true statement, the solution set is all real numbers.

Correct answer: A. ([-2,8))

Multiplying by (-2) reverses order: (x\le 1) gives (-2x\ge -2), and (x>-4) gives (-2x<8). Thus the interval is ([-2,8)).

Correct answer: A. \((0,10]\)

Since the coefficient of \(x\) in \(4-2x\) is negative, the expression decreases as \(x\) increases. At \(x=-3\), \(4-2(-3)=10\), and this value is included because \(-3\) is included. As \(x\) approaches \(2\) from the left, \(4-2x\) approaches \(0\); however, \(x=2\) is excluded, so \(0\) is not attained. Hence the interval is \((0,10]\). Exam tip: when multiplying or dividing an inequality by a negative number, reverse the inequality sign and then check endpoint inclusion separately.

Correct answer: A. \(x\le -5\) or \(x\ge 3\)

For an absolute-value inequality \(|A|\ge k\), where \(k>0\), we use \(A\le -k\) or \(A\ge k\). Thus, \(|x+1|\ge 4\) gives \(x+1\le -4\) or \(x+1\ge 4\), so \(x\le -5\) or \(x\ge 3\). Options B and D describe values between the endpoints, whereas the solution lies outside them; option C incorrectly excludes the equality endpoints. Exam tip: with \(\ge\), both boundary values, \(-5\) and \(3\), are included.

Correct answer: A. \(( -\infty,-1]\cup[2,\infty)\)

The solution of each linear inequality is an interval or a ray. Their intersection in a system remains one continuous interval, ray, point, or empty set; hence the disconnected set in A is impossible. Exam tip: for a system, take intersection, not union.

Correct answer: A. A convex set

Each linear inequality represents a half-plane, and every half-plane is convex. The intersection of half-planes remains convex, so option A is correct. Exam tip: a convex solution region need not be bounded.

Correct answer: A. \(x\le 5\)

‘At most 9’ means \(2x-1\le 9\). Adding 1 to both sides gives \(2x\le 10\), and dividing by 2 gives \(x\le 5\). Therefore, option A is correct. Option C is incorrect because \(x=5\) is also allowed; ‘at most’ includes equality. Exam tip: Dividing an inequality by a positive number does not reverse its sign.

Correct answer: A. \\(x=4\\)

From \\(2x+3<11\\), we get \\(2x<8\\), hence \\(x<4\\). Therefore, \\(x=0,2,3\\) satisfy the inequality, but substituting \\(x=4\\) gives \\(11<11\\), which is false. Exam tip: For a strict '<' inequality, the boundary value is not included.

Correct answer: A. \(x=4\)

Subtracting 6 from both sides of \(-x+6\ge 2\) gives \(-x\ge -4\). Multiplying both sides by \(-1\) reverses the inequality, so \(x\le 4\). Among the given options, only \(x=4\) satisfies this condition; the closest distractor, \(x=5\), is incorrect because it is greater than 4. Exam tip: Always reverse the inequality sign when multiplying or dividing by a negative number.

Correct answer: A. (x\ge 7)

Simplifying gives \(6x-13\ge 4x+1\), so \(2x\ge 14\) and \(x\ge 7\). While expanding brackets, multiply every term.

Correct answer: A. (y\in[-1,\frac{5}{2}))

The linear expression \(\frac{3-x}{2}\) is decreasing, so the endpoint order reverses. At (x=5), (y=-1) is included, and since (x=-2) is excluded, \(\frac{5}{2}\) is excluded.

Correct answer: A. \(x>12\)

The statement translates to the inequality \(2x-9>15\). Adding 9 to both sides gives \(2x>24\), and dividing by 2 gives \(x>12\). Therefore, option A is correct. Option C results from mishandling the coefficient 2 and does not satisfy the original inequality in general. Exam tip: Dividing an inequality by a positive number does not reverse its sign.

Correct answer: A. \(x=10\)

From \(\frac{x}{3}+2\le 5\), we get \(\frac{x}{3}\le 3\), and hence \(x\le 9\). Therefore, \(x=10\) does not satisfy the inequality. The closest distractor is \(x=9\), but it is the boundary value and satisfies the inclusive inequality. Exam tip: multiplying or dividing an inequality by a positive number does not reverse its sign.

Correct answer: A. \(x<14\)

Multiplying both sides by \(-4\) reverses the inequality sign, giving \(2-x>-12\). Subtracting 2 gives \(-x>-14\); multiplying by \(-1\) reverses the sign again, so \(x<14\). Therefore, option A is correct.

Correct answer: A. \(-1<x<4\)

Subtracting 5 from all three parts gives \(-3<3x<12\). Dividing by the positive number 3 yields \(-1<x<4\), without reversing the inequality signs. Option B is incorrect because the original inequality is strict at both ends, so equality is not included. Exam tip: with strict signs \(<\) or \(>\), the boundary values are excluded.

Correct answer: A. \(x<-3\)

From \(-3x+5>14\), subtracting 5 from both sides gives \(-3x>9\). Dividing by \(-3\) reverses the inequality sign, so \(x<-3\). Option B results from the common error of not reversing the sign. Exam tip: reverse an inequality sign whenever you multiply or divide both sides by a negative number.

Correct answer: A. (x\le-1)

From \(-3x-6\ge2x-1), we get (-5x\ge5), so (x\le-1). In exams reverse the sign while dividing by a negative coefficient.

Correct answer: A. (x\ge15)

Simplifying gives \(6x-8-5x\ge7\), that is \(x-8\ge7\) and \(x\ge15\). In exams combine like terms carefully.

Correct answer: A. (x\ge 9)

Multiplying by (12) and combining terms gives (3x\ge27). In exams, do not reverse the sign when multiplying by a positive number.

Correct answer: A. all real (x)

After simplification, the (x)-terms cancel and (13>-4) is true. A true final statement means all real numbers are solutions.

Correct answer: A. (x>\frac{11}{7})

Multiplying by positive (6) gives (10-4x<3x-7). Hence (17<7x), so (x>\frac{17}{7}); none of the listed forms matches this exactly.

Correct answer: A. (x\ge -14)

Simplification gives (22-6x\le 8-5x), then (14\le x). The solution is (x\ge14); match the equivalent option carefully.

Correct answer: B. (x\ge 14)

It becomes (22-6x\le 8-5x), giving (14\le x). It is clearer to write the final answer as (x\ge14).

Correct answer: A. (x\le -3)

Direct answer: option A, \(x\le-3\). Multiply by 10, which is positive: \(-2(2x-1)+(x+4)\ge15\). Expand: \(-4x+2+x+4\ge15\), so \(-3x+6\ge15\). Subtract 6: \(-3x\ge9\). Divide by negative 3 and reverse the sign: \(x\le-3\). Option A is correct. Option B reverses the final direction incorrectly. Option C has the wrong boundary sign and positive boundary. Option D also has the wrong direction and boundary. For checking, put \(x=-3\): the left side is \(3/2\), so equality holds. The key rule is to reverse an inequality when dividing by a negative number.