Introduction to inequalities

Adding the same number to both sides does not change inequality direction. The sign changes only when multiplying or dividing by a negative number.

We get (-4x\ge 12), and dividing by (-4) reverses the sign to (x\le -3). A negative coefficient is the most common trap.

The same (x) cannot be less than (-1) and greater than (3) at once. For compound inequalities, find the common part.

The inequality (x\le -2) includes (-2) and all smaller real values. Infinity always uses an open parenthesis.

Being less than or equal in both directions means equality. Recognizing this helps in comparison questions.

The direct answer is A: \(3x+4>10\). Start with \(x>2\). Multiplying both sides by the positive number 3 keeps the inequality direction: \(3x>6\). Adding 4 to both sides gives \(3x+4>10\). The inequality is strict because \(x\) is strictly greater than 2; it cannot equal 2. Option A is correct. Option B, \(3x+4\ge10\), is weaker and is not the exact conclusion: equality would require \(x=2\), which is forbidden. Option C reverses the relationship and is false; for example, if \(x=3\), the expression is 13, not less than 10. Option D says the expression always equals 10, but its value changes with \(x\). Remember: multiplying or dividing by a positive number keeps the sign; adding or subtracting also keeps it. Only multiplication or division by a negative number reverses an inequality.

Multiplying (x\le -1) by (-5) gives (-5x\ge 5), then adding (2) gives (-5x+2\ge 7). Negative multiplication reverses the sign.

Multiplying both sides by positive (2) gives (x-3\ge 8), so (x\ge 11). A positive denominator does not change the sign.

Multiplying by (-3) reverses the sign, so (5-2x>-21), then (-2x>-26), and dividing by (-2) gives (x<13). The listed choices do not match this result.

Multiplying by (-2) gives (7-3x\ge -10), then (-3x\ge -17), so (x\le \frac{17}{3}). The correct solution is not among the options, so this question is invalid.

Inequalities in opposite directions do not give a fixed order when added directly. The directions should match before adding.

The direct answer is A: \([0,5)\). The interval \(x\in[2,7)\) means \(2\le x<7\). Subtract 2 from every part: \(2-2\le x-2<7-2\), so \(0\le x-2<5\). Therefore the answer is \([0,5)\). The left endpoint is closed because \(x=2\) is allowed, and then \(x-2=0\). The right endpoint is open because \(x=7\) is not allowed, so 5 cannot be included. Option A has exactly these endpoints. Option B, \((0,5]\), wrongly excludes 0 and includes 5. Option C, \([0,5]\), wrongly includes 5. Option D, \((0,5)\), wrongly excludes both valid endpoint 0 and the rest of the correct boundary information. A translation of an interval changes its numerical endpoints but does not change an open bracket into a closed bracket. Memory cue: subtract the same number from both ends, then copy the bracket types.

The compound inequality \(-1<x\le 6\) means two conditions: (i) \(x>-1\) and (ii) \(x\le 6\). Both must be true at the same time, so the equivalent statement is \(x>-1\) and \(x\le 6\).

For \(|x|\ge 3\), we have \(x\le -3\) or \(x\ge 3\). Converting to intervals gives \(( -\infty,-3]\cup[3,\infty)\).

When (x>1), (x) is positive and greater than (1), so its square is also greater than (1). Check the sign condition before squaring.

The negation of \(x<-2\) includes every value that is not less than \(-2\), so it is \(x\ge -2\). Equality must be included when negating a strict inequality. Option C is incorrect because it includes values less than or equal to \(-2\), rather than the values outside \(x<-2\). Exam tip: the negation of \(x<a\) is \(x\ge a\), while the negation of \(x>a\) is \(x\le a\).

From (x\le y) and (y<z), we get (x<z) because one link is strict. In a mixed chain, a strict sign can make the result strict.

Multiplying by a positive number does not change inequality direction. Check the sign of the multiplier before writing the conclusion.

Multiplying (-3\le x<4) by (-3) gives (-12<-3x\le 9), then adding (2) gives (-10<2-3x\le 11). Negative multiplication reverses the order.

At (x=\frac{1}{2}), (x^2=\frac{1}{4}), which is less than (\frac{1}{2}). One counterexample is enough to disprove a universal statement.

\(x^2\ge 0\) for all real \(x\). So \(x^2+1\ge 1>0\). Hence \(x^2+1>0\) is always true, for every real \(x\).

When (x>1), (\frac{1}{x}<1), so (x>\frac{1}{x}). Positivity is important in reciprocal comparisons.

The direct answer is A: both the assertion and reason are true, and the reason correctly explains the assertion. Start with a<b. Adding the same number c to both sides gives a+c<b+c. This operation shifts both numbers equally on the number line, so their order does not change. No restriction is needed on c; it may be positive, zero, or negative. A is correct because the reason states exactly the rule used in the proof. B is wrong because the reason is not merely true separately—it directly justifies the assertion. C is wrong because the reason is true, not false. D is wrong because the assertion is true, not false. Do not confuse addition with multiplying or dividing by a negative number: adding the same quantity preserves the inequality direction. Exam cue: add or subtract the same number on both sides, and the sign stays unchanged.

Multiplying (x>3) by (2) gives (2x>6), and subtracting (5) gives (2x-5>1). Positive multiplication does not reverse the sign.

By the transitive property of inequalities, \(a\le b\) and \(b<c\) imply \(a<c\). Even when \(a=b\), we have \(a=b<c\), so the final inequality remains strict. Equality in option B is not necessary, while options C and D contradict the given chain. Exam tip: in a chain of mixed inequalities, one strict comparison is enough to make the comparison between the outer terms strict.