Introduction to inequalities

“Greater than 12” means that \(s\) must be larger than 12 and cannot equal 12, so \(s>12\) is correct. Option D, \(s\ge 12\), also includes the value 12. Exam tip: use \(>\) for “more than” and \(\ge\) for “more than or equal to.”

At least (20) means (n\ge 20), and at most (45) means (n\le 45). Combining both gives (20\le n\le 45).

First add (7) to both sides to get (5x<25), then divide by (5). Division by a positive number does not change the sign.

First divide by (2) to get (x+3\ge 7), then (x\ge 4). Keep the order correct in bracket questions.

The correct answer is option B: 5 values. The condition \(1<x\le6\) has two parts. The sign \(1<x\) means x must be strictly greater than 1, so 1 is excluded. The sign \(x\le6\) means x may equal 6, so 6 is included. The integers between these limits are 2, 3, 4, 5, and 6. Counting them gives 5 values. Option A, 4, misses one allowed integer. Option B is correct because it lists all five possible integers. Option C, 6, would incorrectly include 1 or another disallowed value. Option D, 7, counts too many values. On a number line, use an open point at 1, a closed point at 6, and consider only integer points between them. Memory cue: a strict sign excludes its endpoint; \(\le\) includes its endpoint.

The governing concept is the degree of the variable. A linear inequality has the variable only to the first power, with no products of variables, roots involving the variable, or higher powers. Options A, B, and D are linear because x has degree one. In option C, the term x² has degree two, so x² − 4 < 0 is a quadratic, not a linear, inequality. Therefore option C is correct. The numerical constants and inequality signs do not determine linearity by themselves; the exponent of the variable does. Even though option C can be solved as an inequality, its form is nonlinear. This distinction is important before selecting an algebraic or graphical method.

The symbol ([-4) shows that (-4) is included and (2)) shows that (2) is not included. Therefore (-4\le x<2) is correct.

The governing concept is translating verbal comparison into mathematical symbols. Let the unknown number be x. Its twice value is 2x. The phrase “at least 18” means 18 or any number greater than 18, so equality must be included. Its correct symbol is ≥, giving 2x ≥ 18; therefore option B is correct. Option A, 2x > 18, would mean strictly greater than 18 and would wrongly exclude the allowed boundary value 2x = 18. Options C and D describe values less than 18 or at most 18, which contradict the phrase “at least.” For example, if x = 9, then 2x = 18 and the original sentence is true, confirming why the equality sign is essential.

From (4-x\le 0), we get (-x\le -4), and dividing by a negative gives (x\ge 4). Reverse the sign when removing (-x).

The governing concept is the meaning of the inclusive inequality symbol ≥. It means “greater than or equal to,” so equality at the boundary is allowed. Substituting y = 0 into y ≥ 0 gives 0 ≥ 0, which is true because both sides are equal. Therefore option A is correct. Option B would be correct for a strict statement such as y > 0, where zero is excluded. Options C and D are unsuitable because the value of y is explicitly given and the statement can be evaluated directly. On a number line, the solution of y ≥ 0 begins at zero with a closed point and extends to the right, again showing that zero belongs to the solution set.

“At least 50” means 50 or more, so m may be equal to 50 or greater than 50. Hence, the correct inequality is \(m\ge 50\). Option A, \(m>50\), is incorrect because it excludes m=50. Exam tip: use \(\ge\) for “at least” and \(\le\) for “at most.”

“Not more than 75” means that c can be less than or equal to 75, so the correct inequality is \(c\le 75\). Option A, \(c<75\), wrongly excludes the value 75. Exam tip: translate “not more than” or “at most” as \(\le\).

The governing property is that multiplication by a positive number preserves the order in an inequality. Starting with p > 6, multiply both sides by 3: 3p > 3×6 = 18. Hence option A is correct. The direction remains greater than because 3 is positive. Option B incorrectly reverses the sign, which would happen only with multiplication by a negative number. Option C omits the multiplication from the left side, while option D changes both the operation and direction. A quick check with p = 7 confirms the result: 7 > 6 and 3×7 = 21 > 18. Thus every p greater than 6 produces a value 3p greater than 18.

A linear inequality is solved much like an equation, but the effect of an operation on the inequality sign must be remembered. Dividing or multiplying both sides by a positive number keeps the inequality direction unchanged; only multiplication or division by a negative number reverses it. Starting with 4x≥20, divide both sides by the positive number 4: 4x/4≥20/4, so x≥5. Thus every number 5 or greater satisfies the original inequality, while a number below 5 does not. Substitution confirms the boundary: x=5 makes both sides equal to 20. Therefore option A is correct. Options B and D reverse the direction without cause, while C incorrectly subtracts 4 rather than dividing by 4.

In (x>2), (2) is not included and greater numbers lie to the right. So use an open circle and shade right.

The sign (\le) includes the boundary, and smaller numbers lie to the left. So use a closed circle at (-4) and shade left.

The interval \(( -1, \infty )\) means all real numbers greater than \(-1\), not including \(-1\). Therefore the inequality is \(x>-1\).

In (r\ge -2), (-2) and greater numbers are included. With an equality sign, the boundary value is also a solution.

“45 or more” means that the marks may be exactly 45 or greater than 45. Therefore, the correct inequality is \(s\ge 45\). Option A is incorrect because \(s>45\) excludes 45. In exams, translate “or more” or “at least” as \(\ge\).

“Not more than 12 kilograms” means that the weight can be equal to 12 kilograms or less than 12 kilograms. Therefore, the correct inequality is \(w\le 12\). Option A is incorrect because \(w<12\) excludes a weight of exactly 12 kilograms. Exam tip: phrases such as “maximum,” “not more than,” and “at most” indicate \(\le\).

“Less than 30 rupees” means that the price cannot be equal to 30, so \(p<30\) is correct. Option B includes the value 30, which is not allowed by the phrase “less than.” In exams, remember that “less than” uses \(<\), while “less than or equal to” uses \(\le\).

‘At least 5 years’ means 5 years or more, so the correct inequality is \(a\ge 5\). Option B is incorrect because \(a\le 5\) means 5 years or less. In exams, remember that ‘at least’ corresponds to \(\ge\), while ‘at most’ corresponds to \(\le\).

Substituting \(x=-1\) in option B gives \(-1\le -1\), which is true because equality is included. Option A gives the false statement \(-1<-3\), and options C and D are also false for \(x=-1\). In an exam, substitute the given value directly and check carefully whether equality is included in the inequality sign.

Start with the inequality \\(9-x<4\\). Subtract 9 from both sides to obtain \\(-x<-5\\). To isolate x, multiply or divide both sides by \\(-1\\). Multiplication by a negative number reverses the inequality sign, so the result becomes \\(x>5\\). Therefore every real number greater than 5 satisfies the original inequality, and option A is correct. The sign reversal is essential; leaving the sign as < would give an incorrect solution.

A quick check confirms the result. If \\(x=6\\), then \\(9-6=3<4\\), so it works. If \\(x=5\\), then \\(9-5=4\\), which is not strictly less than 4, so 5 is excluded. Values below 5 fail the condition, while all values above 5 work. Thus the solution set is \\( (5,\infty)\\), equivalently \\(x>5\\), exactly as stated in the supplied answer A.

In (x>-3), (-3) is not included and greater numbers lie to the right. Use an open circle on the number line.