Introduction to inequalities

The governing concept is interpretation of a compound inequality. The expression 0 ≤ x < 10 contains two conditions that must hold together: x must be at least 0, and x must be strictly less than 10. The symbol ≤ at the left endpoint includes 0, whereas the symbol < at the right endpoint excludes 10. Therefore the solution interval is [0, 10), and option A states this correctly in words. For example, 0, 4.5, and 9 are included, but 10 is not included. Option B contradicts both conditions, option C includes only one value instead of an interval, and option D describes values below 0, which are outside the required range.

If the minimum value of n is 18, then n can be equal to 18 or any value greater than 18. Hence, \(n\ge 18\) is correct. Options A and B describe an upper limit of 18, while option D excludes 18 itself. Exam tip: use \(\ge\) for a minimum value and \(\le\) for a maximum value.

‘Less than or equal to 25’ means that \(u\) may be smaller than 25 or exactly equal to 25. This is represented by \(\le\), so \(u\le 25\) is correct. Option A excludes the case \(u=25\). Exam tip: when ‘or equal to’ is stated, choose the inequality symbol that includes the equality bar.

The governing concept is the addition property of inequalities: adding the same real number to both sides preserves the original order. Since a > b, adding 5 to the left side and 5 to the right side gives a + 5 > b + 5. The inequality sign does not reverse because reversal occurs only when both sides are multiplied or divided by a negative number. Option B contradicts the given order, while option C is not obtained from the stated operation and is generally false. Option D changes only one side by multiplication and is not guaranteed. Therefore, option A is the only statement that must always be true.

An inequality preserves its direction when both sides are multiplied by a positive number. Since \(a<b\) and \(k>0\), multiplying each side by k gives \(ak<bk\). The positivity of k is essential: multiplication by a negative number would reverse the sign, while multiplication by zero would destroy the strict comparison.

Therefore option A, \(ak<bk\), follows directly from the order rule for positive multiplication. Adding k to both sides would also preserve direction and give \(a+k<b+k\), but that is not what option C states. Similarly, subtracting k gives \(a-k<b-k\), not the reversed inequality in option D. Thus A is the correct choice.

When both sides of an inequality are multiplied by a negative number \(k\), the inequality sign reverses. Thus, from \(a>b\), we get \(ak<bk\), so option B is correct. Option D is also incorrect because division by a negative number reverses the sign, giving \(\frac{a}{k}<\frac{b}{k}\). Exam tip: Reverse the inequality sign whenever you multiply or divide both sides by a negative number.

The inequality is solved by undoing the division by 6. Since 6 is positive, multiplying both sides by 6 does not reverse the inequality sign. This is the same basic rule used for equations, with the additional need to watch the sign of any multiplier or divisor.

Starting from \(x/6\le2\), multiply both sides by 6: \(6(x/6)\le6(2)\). The left side becomes x and the right side becomes 12, so the result is \(x\le12\). Therefore option A is correct. The boundary value 12 is included because the original sign is \(\le\); values greater than 12 do not satisfy the inequality.

At least (12) means (g\ge 12), and at most (28) means (g\le 28). Combining both gives (12\le g\le 28).

First add (5) to both sides to get (6x\ge 24), then divide by (6). Division by a positive number does not change the sign.

The correct answer is option A: \(x<7\). Begin with \(3(x-2)<15\). Divide both sides by 3, a positive number, so the inequality sign does not change: \(x-2<5\). Add 2 to both sides: \(x<7\). Option A is exactly this result. Option B, \(x>7\), reverses the inequality without justification. Option C, \(x<5\), forgets to add 2 after obtaining \(x-2<5\). Option D, \(x>5\), has both the wrong direction and the wrong boundary. Checking with x=6 gives \(3(6-2)=12<15\), so values below 7 work; x=7 gives 15<15, which is false, so 7 is excluded. Memory cue: dividing by a positive number keeps the sign; adding or subtracting moves the constant normally.

The possible integers are (2,3,4,5,6,7), so there are (6) values. The sign (\le) includes the endpoint but (<) does not.

The symbol ((-3) shows that (-3) is not included and (5]) shows that (5) is included. Therefore (-3<x\le 5) is correct.

The governing concept is translating verbal statements into algebraic inequalities. Let the unknown number be x. The phrase “adding 4 to a number” means x + 4, not x − 4 or 4x. The phrase “greater than 10” is represented by the strict symbol >, so the complete inequality is x + 4 > 10. Therefore option A is correct. The statement does not say “at least 10,” which would require ≥, nor does it say “less than 10,” which would require <. Option B reverses the comparison, option C subtracts 4 instead of adding it, and option D treats addition as multiplication. Careful attention to both the operation and comparison phrase gives the unique answer.

From (7-x\ge 2), we get (-x\ge -5), and dividing by a negative gives (x\le 5). Reverse the sign when removing (-x).

The governing concept is reversal of the inequality sign under multiplication by a negative number. Starting with x ≥ −3, multiply both sides by −2. Because −2 is negative, the direction changes from ≥ to ≤. The right side becomes (−3)(−2) = 6, so the result is −2x ≤ 6. Therefore option B is correct. Option A keeps the original direction and is the standard sign-reversal error. Option C changes the coefficient to positive 2 even though the multiplier is −2. Option D correctly has a negative coefficient and a reversed sign but calculates the right-hand product incorrectly. A test value such as x = 0 satisfies the original inequality and gives 0 ≤ 6, confirming the result.

First divide by (2) to get (3-x\le 4), then (-x\le 1) gives (x\ge -1). Remember to reverse the sign when removing a negative variable.

The governing concept is substitution into each inequality, followed by a direct comparison. Put x = 2 into option A: 3(2) − 1 = 5, and 5 > 7 is false. In option B, 4(2) + 2 = 10, so 10 ≤ 9 is false. In option C, 5 − 2 = 3, and 3 ≥ 3 is true because a greater-than-or-equal relation includes equality. In option D, 2(2) + 6 = 10, so 10 < 9 is false. Therefore option C is the only inequality satisfied by x = 2. This question also tests the distinction between strict symbols > or < and inclusive symbols ≥ or ≤; ignoring equality would incorrectly reject option C.

The governing concept is verifying an inequality through substitution while recognizing whether equality is allowed. Set x = 0 in every option. A becomes 2(0) + 1 = 1, so 1 < 0 is false. B becomes 4 − 0 = 4, and 4 ≤ 3 is false. C becomes −0 + 2 = 2, giving 2 ≥ 2, which is true because ≥ includes equality. D becomes 0 − 5 = −5, and −5 > 1 is false. Hence option C is the only correct answer. The equality case is decisive: the numerical expressions on the two sides of C are equal, and the symbol permits that equality. If C had used > instead of ≥, it would not have been satisfied.

The governing idea is testing membership in a closed interval. The statement 2 ≤ x ≤ 6 means that x may be any real number from 2 through 6, and both endpoints are included because the symbols are ‘less than or equal to.’ Test the choices directly: x = 6 satisfies 2 ≤ 6 and 6 ≤ 6, so it belongs to the solution set. The value 7 is greater than the upper limit, while 1 and 0 are less than the lower limit. Therefore, option A is the only valid solution. A strict inequality would exclude an endpoint, but that is not the case here.

The governing concept is translating verbal comparison into an inequality symbol. Let x represent the number. The phrase “greater than 6” means that x is strictly larger than 6, so it is written as x>6. The word “than” sets 6 as the comparison value, and “greater” points to the right side of the number line. Option A is therefore correct. The symbol < in option B means less than, so it reverses the comparison. Option C means greater than or equal to 6 and would include 6, but the statement does not say “at least 6.” Option D means less than or equal to 6. Since equality is not mentioned, the strict sign > is required.

The governing concept is the distinction between strict and inclusive inequalities. In x<−2, the symbol < means that x must be smaller than −2, but it cannot equal −2. Therefore the boundary value −2 is excluded, making option A correct. On a number line this is shown with an open circle at −2 and shading to the left. Option B would correspond to an inequality such as x≤−2, where equality is allowed. Option C incorrectly treats the boundary as the only solution, while option D selects values greater than −2, which is the opposite direction. For example, −3 satisfies the inequality, but −2 does not because −2<−2 is false.

The governing concept is the sign rule for multiplication and the meaning of a strict inequality. Since x > 0, x is positive, while −2 is negative. The product of a negative number and a positive number is negative, so −2x < 0. The inequality remains strict because x cannot equal zero; consequently, −2x cannot equal zero either. Option B is false because the product is not positive. Option C is impossible because it would require x = 0. Option D is also false: a negative number is not greater than or equal to zero. Therefore option A is the only statement that follows for every positive value of x.

The governing property is preservation of order under multiplication. If a>b and c is positive, multiplying both sides by c keeps the direction unchanged, so ac>bc. For example, if a=5, b=3, and c=2, then 10>6 remains true. If c is negative, multiplication reverses the order and produces ac<bc, not the expression shown. If c=0, both products become equal to zero, so strict inequality is lost. Merely stating c≠0 is insufficient because c could be negative and would reverse the sign. Therefore the condition that always gives ac>bc is c>0, represented by option A.

The governing concept is solving a strict linear inequality and then applying the integer restriction. Start with 4x + 9 < 1. Subtracting 9 gives 4x < −8. Since division by the positive number 4 does not reverse the inequality, x < −2. Because x must be an integer, the greatest integer strictly less than −2 is −3. Thus option D is correct. The value −2 is not permitted because it produces equality: 4(−2) + 9 = 1, whereas the condition requires a value less than 1. Values 0 and −1 are greater than −2 and clearly fail the inequality. The remaining acceptable integers are −3, −4, −5 and so on, with −3 the greatest.

A compound inequality must be read as two conditions at the same time: x is greater than −4 and x is less than or equal to 3. The strict sign < excludes the left endpoint −4, while the inclusive sign ≤ includes the right endpoint 3. Therefore −3 and 0 are inside the interval, and 3 is also valid because equality at the upper endpoint is allowed. However, x = −4 is exactly equal to the excluded lower boundary, so it is not a solution. Option D is correct. On a number line, the solution set is shown with an open circle at −4, a closed circle at 3, and all points between them selected. This endpoint test distinguishes D from the other three choices without any additional calculation.