Algebraic Solution of Linear Inequalities in One Variable

The LCM of the denominators is 6. Multiplying the inequality by 6 gives \(2(2x-5)+(x+1)\le 24\). Thus, \(5x-9\le 24\), so \(5x\le 33\) and \(x\le \frac{33}{5}\). Option C wrongly excludes equality, although the original inequality contains \(\le\). Exam tip: multiplying an inequality by a positive number does not reverse its sign.

On simplifying, \(5x-5-6x-8<9\) becomes \(-x-13<9\). Adding 13 to both sides gives \(-x<22\). Multiplying by \(-1\) reverses the inequality sign, so \(x>-22\). The option \(x<-22\) results from forgetting to reverse the sign when multiplying by a negative number. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative quantity.

The LCM of the denominators 2 and 3 is 6. Multiplying both sides of the inequality by the positive number 6 gives \(9-12x\ge 30-2x\). Hence, \(-21\ge 10x\), so \(x\le -\frac{21}{10}\). Option C is close, but the original inequality has \(\ge\), so the boundary value \(-\frac{21}{10}\) is included. Exam tip: multiplying or dividing an inequality by a positive number does not reverse its sign.

Multiplying both sides by the positive number 4 does not change the inequality sign: \(8-(5x-1)<2x+14\). Thus, \(9-5x<2x+14\), which gives \(-7x<5\). Dividing by \(-7\) reverses the inequality, so \(x>-\frac{5}{7}\). Option C incorrectly includes equality, whereas the original inequality is strict. Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

In \(2<x<7\), both inequalities are strict, so neither 2 nor 7 is included in the solution set. Hence, the solution set is the open interval \((2,7)\). Options B and C represent half-open intervals, while option D is a closed interval because both endpoints are included. Exam tip: An endpoint is open for < or >, and closed for \(\le\) or \(\ge\).

Simplifying the left-hand side gives \(-2(5-x)+3(x-4)=-10+2x+3x-12=5x-22\). Thus, \(5x-22>6x+1\), so \(-x>23\). On multiplying both sides by \(-1\), the inequality sign reverses, giving \(x<-23\). \(x\le -23\) is incorrect because the original inequality is strict, \(>\). Exam tip: always reverse an inequality sign when multiplying or dividing by a negative number.

Write \(0.75x\) as \(\frac{3x}{4}\). Adding \(\frac{x}{4}\) to both sides gives \(x+\frac{1}{2}<2\). Subtracting \(\frac{1}{2}\) gives \(x<\frac{3}{2}\), so option A is correct. Option C is incorrect because the inequality is strict: at \(x=\frac{3}{2}\), both sides are equal, which is not allowed by \(<\). Exam tip: Reverse an inequality sign only when multiplying or dividing both sides by a negative number.

Multiplying both sides by the positive number \(5\) does not reverse the inequality: \(9x+4\ge 10x-15\). Rearranging gives \(19\ge x\), or \(x\le 19\). The closest distractor, \(x<19\), is incorrect because \(x=19\) also satisfies the inequality. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Multiplying by (14) gives (42-2(2x+5)\le 7(1-x)). This gives (32-4x\le 7-7x), so (3x\le -25).

On expanding the bracket, \(8-3(2-x)=8-6+3x=2+3x\). Thus, \(2+3x\ge 4x-1\), which gives \(3\ge x\), or \(x\le 3\). The closest distractor, \(x<3\), is incorrect because at \(x=3\) both sides are equal, and \(\ge\) includes equality. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

Using the LCM 6, the left-hand side becomes \(\frac{2x-(x-5)}{6}=\frac{x+5}{6}\). Thus, \(\frac{x+5}{6}>2\) gives \(x+5>12\), so \(x>7\). The option \(x\geq 7\) is not correct because the original inequality is strict; at \(x=7\), the left-hand side equals 2. Exam tip: multiplying an inequality by a positive number does not reverse its sign.

The inequality is \(\frac{-4x+1}{3}\le 5-x\). Multiplying both sides by the positive number \(3\) does not reverse the inequality sign: \(-4x+1\le 15-3x\). Adding \(4x\) to both sides gives \(1\le 15+x\), so \(-14\le x\), i.e. \(x\ge -14\). Hence, option A is correct. \(x>-14\) is incorrect because \(x=-14\) also satisfies the inequality. Exam tip: reverse an inequality sign only when multiplying or dividing by a negative number.

Given \(2(1-3x)<4-7x\), expanding the bracket gives \(2-6x<4-7x\). Adding \(7x\) to both sides gives \(2+x<4\), so \(x<2\). Hence, \(x<2\) is correct. \(x\leq2\) is not correct because the original inequality is strict, so \(x=2\) is excluded. Exam tip: Reverse an inequality sign only when multiplying or dividing both sides by a negative number.

Multiplying both sides by the positive number 4 does not change the inequality sign: \(20-2(x-3)\ge 3x+1\). Simplifying gives \(26-2x\ge 3x+1\), or \(25\ge 5x\). Dividing by 5 gives \(x\le 5\). The close distractor \(x<5\) is incorrect because \(x=5\) also satisfies the original inequality. Exam tip: the inequality sign remains unchanged when multiplying or dividing by a positive number.

Expanding the right-hand side gives \(3(2x+1)=6x+3\). Thus, the inequality becomes \(6x+2<6x+3\). Subtracting \(6x\) from both sides gives \(2<3\), which is always true. Therefore, the inequality is satisfied by every real value of \(x\). Conditions such as \(x<1\) or \(x>1\) do not arise because the variable terms cancel completely. Exam tip: first expand brackets, then cancel like variable terms before deciding the solution set.

An inequality compares two expressions and includes all values that make the comparison true. The sign 
m less than or equal to
m means that equality is also allowed, so the boundary value may be included. The important rule here is that adding, subtracting, or dividing by a positive number does not reverse the inequality sign. Only multiplication or division by a negative number would reverse it.

Start with 
m 7x-4
aisebox{0.1em}{\le}2x+16
m. Subtract 
m 2x
m from both sides to obtain 
m 5x-4
aisebox{0.1em}{\le}16
m. Add 4 to both sides: 
m 5x
aisebox{0.1em}{\le}20
m. Divide by positive 5, giving 
m x
aisebox{0.1em}{\le}4
m. Thus option A is correct; values such as 4 and 0 satisfy the inequality, while a value greater than 4 does not.

Simplifying the right-hand side gives \(3(4-2x)+x=12-6x+x=12-5x\). Thus, the inequality becomes \(12-5x\ge 12-5x\), which is true for every real value of \(x\). The conditions \(x\le0\) and \(x\ge0\) include only some values, whereas this inequality has no restriction. Exam tip: Simplify both sides first; if they become identical with \(\ge\) or \(\le\), the solution is all real numbers.

Multiplying both sides by the positive number \(6\) does not change the inequality sign: \(24-9x>6+x\). Hence, \(18>10x\), so \(x<\frac{9}{5}\). The option \(x\leq\frac{9}{5}\) is incorrect because the original inequality is strict; at \(x=\frac{9}{5}\), both sides become equal. Exam tip: Reverse an inequality sign only when multiplying or dividing by a negative number.

The correct answer is option A: x ≥ 55/27. Begin with 2.2x − 1.1 ≥ 4.4 − 0.5x. Add 0.5x to both sides: 2.7x − 1.1 ≥ 4.4. Add 1.1: 2.7x ≥ 5.5. Write decimals as fractions: 2.7 = 27/10 and 5.5 = 55/10. Dividing by 2.7, which is positive, gives x ≥ 5.5/2.7 = (55/10)/(27/10) = 55/27. Option A matches. Option B reverses the inequality without a negative division. Option C changes ≥ to >, wrongly excluding equality. Option D reverses the direction and also uses a strict sign. Checking x = 55/27 gives equality, so the endpoint must be included. The duplicated slash in the options appears to be formatting, but the intended mathematical answer is clear. Memory cue: only a negative multiplier or divisor reverses an inequality sign.

The left side is (x-13). The inequality (x-13\le x-20) gives false (-13\le -20), so there should be no solution.

The correct solution is obtained by clearing the positive denominators. Multiply both sides by 15, the least common multiple of 5 and 3; because 15 is positive, the inequality direction stays the same: 3(x − 8) ≥ 5(2x + 1) − 60. Expanding gives 3x − 24 ≥ 10x + 5 − 60 = 10x − 55. Add 55 to both sides: 3x + 31 ≥ 10x. Subtract 3x: 31 ≥ 7x. Dividing by positive 7 yields x ≤ 31/7. Thus option A is correct. The supplied original option text showing 47/7 was inconsistent with the algebra, so it has been corrected here to 31/7. Options B and D reverse the direction, while C wrongly changes the inclusive endpoint to a strict one.

On expanding the brackets, \(9-4x-4<6-2x\), so \(5-4x<6-2x\). Adding \(4x\) to both sides gives \(5<6+2x\), hence \(-1<2x\). Dividing by 2 gives \(x>-\frac{1}{2}\). The option \(x\geq-\frac{1}{2}\) is incorrect because the original inequality is strict, so \(x=-\frac{1}{2}\) is not included. Exam tip: dividing by a positive number does not reverse the inequality sign.

Multiplying both sides by the positive number 4 does not change the inequality sign: \(-24+10x\le x-3\). Hence \(9x\le21\), so \(x\le\frac{21}{9}=\frac{7}{3}\). Option B incorrectly excludes \(x=\frac{7}{3}\), which satisfies the inequality. Exam tip: Reverse an inequality only when multiplying or dividing by a negative number.

Expanding the right-hand side gives \(2(2x+5)=4x+10\). Thus, \(4x+7>4x+10\). Subtracting \(4x\) from both sides gives \(7>10\), which is false. Therefore, the inequality is not true for any real value of \(x\), so there is no solution. Options such as \(x<3\) or \(x>3\) are incorrect because the \(x\)-terms cancel completely. Exam tip: When like variable terms cancel, always check the resulting numerical inequality.

Multiplying both sides by 18 does not change the inequality sign because 18 is positive: \(2(2-3x)<3(x+4)\). Thus, \(4-6x<3x+12\), so \(-9x<8\). On dividing by \(-9\), the inequality sign reverses, giving \(x>-\frac{8}{9}\). \(x\geq-\frac{8}{9}\) is incorrect because the original inequality is strict. Exam tip: always reverse the inequality sign when multiplying or dividing by a negative number.