Algebraic Solution of Linear Inequalities in One Variable

Correct answer: A. (p<9)

For the interval to be non-empty, the left endpoint must be less than the right open endpoint. From (2p-5<p+4), (p<9).

Correct answer: B. \(x \geq 14\)

Multiplying both sides by the positive number 10 does not change the inequality sign: \(6x-4-5x-20\geq -10\). Thus, \(x-24\geq -10\), so \(x\geq 14\). The option \(x>14\) is incorrect because it excludes 14, and \(x=14\) satisfies the original inequality. Exam tip: Multiply by the LCM of the denominators to solve fractional inequalities efficiently.

Correct answer: C. ( \frac{1}{3}\leq x<\frac{13}{3} )

Dividing by (-3) reverses both inequality signs. Hence the solution is ( \frac{1}{3}\leq x<\frac{13}{3} ).

Correct answer: D. -2

From \(ax+7<1\), we get \(ax<-6\). For the solution to be \(x>3\), \(a\) must be negative because dividing an inequality by a negative number reverses its sign. At the boundary value \(x=3\), \(3a=-6\), so \(a=-2\). Check: \(-2x+7<1\Rightarrow -2x<-6\Rightarrow x>3\). If \(a=2\), the result is \(x<-3\), so it is not correct. Exam tip: Always reverse the inequality sign when multiplying or dividing by a negative quantity.

Correct answer: A. (5)

The first inequality gives (x<11), and the second gives (x\geq 6). The integers are (6,7,8,9,10), so there are (5) values.

Correct answer: B. (x\in\mathbb{R})

Both sides become identical, so the inequality is true for every real (x). In an identity-type inequality, all real numbers are the solution.

Correct answer: D. (x<-\frac{1}{3})

Simplification gives (-6x>2), so (x<-\frac{1}{3}). Reverse the sign when dividing by a negative coefficient.

Correct answer: C. (x\ge \frac{26}{7})

Clearing denominators gives (7x-14\ge 12), hence (x\ge \frac{26}{7}). For fractions, multiply by the LCM first.

Correct answer: B. (x\ge \frac{15}{2})

Simplification gives (2x\ge 15), so (x\ge \frac{15}{2}). Apply the negative sign carefully while opening brackets.

Correct answer: C. (x\ge 3)

The left side becomes (-5), so (-5\le x-8) and (x\ge 3). When variables cancel, place the remaining constant correctly.

Correct answer: D. \(\varnothing\)

After simplification, (2\ge 4), which is false. A false constant inequality has an empty solution set.

Correct answer: A. (x>2)

Clearing denominators gives (12x+11>35), so (x>2). Multiplying by a positive LCM does not change the sign.

Correct answer: C. (x\in\mathbb{R})

The left side becomes (-\frac{11}{5}), and (-\frac{11}{5}<\frac{5}{2}) is true. Therefore every real (x) is a solution.

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

Subtract 6 from all three parts of the compound inequality to get \(-2\le 2x<8\). Dividing every part by the positive number 2 does not reverse the inequality signs, giving \(-1\le x<4\). Thus, \(-1\) is included but \(4\) is excluded. Option B incorrectly excludes \(-1\). Exam tip: inequality signs remain unchanged when dividing by a positive number, but reverse when dividing by a negative number.

Correct answer: C. \(-23<x\le 1\)

Since 3 is positive, multiplying all three parts of the compound inequality by 3 does not reverse any inequality sign: \(-21<x+2\le 3\). Subtracting 2 throughout gives \(-23<x\le 1\). Hence, option C is correct. In option D, the open and closed endpoints have been reversed. Exam tip: An inequality sign reverses only when multiplying or dividing by a negative number.

Correct answer: B. \(\varnothing\)

Multiplying the first inequality by 2 gives \(3x-1\ge 2x+8\), so \(x\ge 9\). From the second inequality, \(2x<12\), hence \(x<6\). No number can satisfy both \(x\ge 9\) and \(x<6\), so the combined solution is \(\varnothing\). The option \(x\ge 9\) solves only the first inequality, not both. Exam tip: for “and”, take the intersection of the solution sets.

Correct answer: A. (x<13)

The first inequality gives (x<13), and the second gives (x\le -3). The second set is contained in the first, so the answer is (x<13).

Correct answer: D. \(x \ge 3\)

Expanding gives \(5-10x\) on the left and \(3-8x-4=-1-8x\) on the right. Thus, \(5-10x\le -1-8x\), so \(6\le 2x\). Dividing by 2 does not reverse the inequality, giving \(x\ge 3\). The closest distractor is \(x>3\), but \(x=3\) also satisfies the inequality. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Correct answer: B. \(x>1\)

Multiplying the inequality by 2 gives \(4x+12-(3x-1)>14\). On simplifying, \(x+13>14\), so \(x>1\). The option \(x\geq1\) is not correct because the original inequality is strict, and \(x=1\) does not satisfy it. Exam tip: multiplying an inequality by a positive number does not reverse its sign.

Correct answer: A. \(x\le -16\)

Multiplying both sides by the positive number 10 does not change the inequality sign: \(2(4x+9)\le 5(x-6)\). Thus, \(8x+18\le 5x-30\), so \(3x\le -48\), giving \(x\le -16\). The close distractor \(x<-16\) is incorrect because at \(x=-16\), both sides are equal, so the endpoint must be included. Exam tip: an inequality sign does not reverse when multiplying or dividing by a positive number.

Correct answer: A. \(x\le \frac{7}{2}\)

Simplifying the left-hand side gives \(9-3(x-4)=21-3x\), while the right-hand side gives \(2(7-x)+x=14-x\). Thus, \(21-3x\ge14-x\), or \(7\ge2x\). Since 2 is positive, dividing both sides by 2 does not reverse the inequality sign, giving \(x\le\frac{7}{2}\). The option \(x<\frac{7}{2}\) is incorrect because \(x=\frac{7}{2}\) also satisfies the equality case. Exam tip: An inequality sign remains unchanged when dividing by a positive number.

Correct answer: D. \(x\le \frac{9}{2}\)

Multiplying both sides by 6 gives \(2(2x-5)+(x+1)\le 3x\). Hence, \(5x-9\le 3x\), so \(2x\le 9\) and \(x\le \frac{9}{2}\). Option B excludes \(\frac{9}{2}\), but this endpoint must be included because the inequality has \(\le\). Exam tip: For inequalities involving fractions, first multiply by a positive LCM to remove denominators.

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

In the first inequality, \(7x-2(3x+5)=7x-6x-10=x-10\). Thus, \(x-10\ge4\), so \(x\ge14\). From the second inequality, \(x-1>6\), giving \(x>7\). Since both conditions must hold, take the intersection of the two solution sets: \(x\ge14\). The option \(x>7\) satisfies only the second inequality. Exam tip: for “and,” always take the intersection of the solution sets.

Correct answer: B. \(x\in\mathbb{R}\)

Simplifying the left-hand side gives \(\frac{1}{3}(6x-9)=2x-3\) and \(\frac{1}{2}(4x+2)=2x+1\). Hence, the left-hand side is \((2x-3)-(2x+1)=-4\). Since \(-4\geq -5\) is always true, the inequality holds for every real value of \(x\). The option \(x\geq -5\) restricts \(x\), but this inequality imposes no restriction. Exam tip: When variable terms cancel, check whether the resulting constant statement is true or false.

Correct answer: C. \(x<8\)

Expanding the left side gives \(\frac{3x}{4}-6\), while simplifying the right side gives \(\frac{x}{2}-4\). Thus, \(\frac{3x}{4}-6<\frac{x}{2}-4\), which gives \(\frac{x}{4}<2\), and hence \(x<8\). The option \(x\leq 8\) is not correct because the original inequality is strict; \(x=8\) is not a solution. Exam tip: After moving terms in a fractional inequality, carefully check the coefficient and the inequality sign.