Introduction to inequalities

The negation of \(x<7\) includes every value that is not less than 7, so it is \(x\ge 7\). Option B, \(x>7\), wrongly excludes \(x=7\), which must be included in the negation. Exam tip: interpret ‘not less than’ as ‘at least’.

From (4-3x\le -8), we get (-3x\le -12), and dividing by (-3) gives (x\ge 4). Division by a negative reverses the sign.

Multiplying by positive (5) gives (2x+1>15), so (2x>14) and (x>7). A positive denominator does not change the sign.

Multiply both sides by \(-2\) (a negative number), so the inequality sign reverses:
\[\frac{x-4}{-2}\ge 6\Rightarrow x-4\le 6\cdot(-2)\Rightarrow x-4\le -12\Rightarrow x\le -8.\]

The common part of both conditions starts at (-1) and goes up to before (4). The value (-1) is included and (4) is excluded.

In option A, \(-3<-1\), but squaring gives \((-3)^2=9\) and \((-1)^2=1\), so \(9>1\), not \(9<1\). Hence, it is a counterexample to the stated implication. In options B, C, and D, both numbers are non-negative, so their squares preserve the original order. Exam tip: Check the signs of the numbers before squaring an inequality.

Multiplying (x\le 2) by (-4) gives (-4x\ge -8), and adding (7) gives (7-4x\ge -1). Negative multiplication reverses the sign.

From (1<x<4), we get (0<x-1<3), then dividing by positive (3) gives (0<\frac{x-1}{3}<1). Division by a positive number does not change the sign.

The direct answer is A: there is no real number satisfying both conditions. The first inequality x≤6 allows 6 and every number smaller than 6. The second inequality x>6 allows only numbers strictly larger than 6. These two sets have no common element, because a number cannot be at most 6 and greater than 6 at the same time. Therefore their intersection is empty, written as ∅, or “no x∈R.” A is correct. B is wrong because x=6 satisfies x≤6 but fails x>6. C is wrong because numbers greater than 6 do not satisfy the first inequality, and numbers less than or equal to 6 do not satisfy the second. D is wrong because x>0 includes values such as 1, which fail x>6, and it still does not describe the common solution. On a number line, x≤6 shades left including 6, while x>6 shades right excluding 6; there is no overlap. Memory cue: opposite strict sides at the same boundary give an empty intersection.

Multiplying (x\ge -2) by (-\frac{1}{2}) gives (-\frac{x}{2}\le 1), then adding (5) gives (-\frac{x}{2}+5\le 6). A negative multiplier reverses the sign.

The direct answer is A: x=0. For every real number x, its square satisfies x²≥0. Therefore x²≤0 can happen only when x² is exactly 0. A square is zero only when its base is zero, so x=0. A is correct. B, x<0, is wrong because a negative number has a positive square; for example, (−2)²=4, not a value at most zero. C, x>0, is wrong for the same reason: a positive number also has a positive square, for example 2²=4. D, all real x, is wrong because nonzero values produce positive squares; x=1 gives x²=1>0. The solution set contains exactly one point, zero. On a graph or number line, mark only 0 with a closed point because equality is included. Do not mistakenly think that “≤0” means all negative x; it describes the square’s value, not x itself. Memory cue: a real square never goes below zero, and it touches zero only at zero.

Subtracting (3) first gives (-10<2x\le 6). Then dividing by positive (2) gives the correct solution (-5<x\le 3).

Since (x>0), multiplying by (x) gives (x^2<1). With the positive condition, the solution is (0<x<1).

From (6-3x<x+5), we get (1<4x), so (x>\frac{1}{4}). While moving variable terms to one side, keep the sign and order carefully.

From (7-2x\le x+1), we get (6\le 3x), so (x\ge 2). Apply the same operation to both sides while simplifying.

From (\frac{3x}{2}<9), we get (3x<18), so (x<6). Dividing by a positive number does not reverse the sign.

Multiplying by positive (6) gives (2(2x-1)\ge 3(x+5)), hence (x\ge 17). A positive multiplier used to clear denominators does not change the sign.

Subtracting (6) gives (-10\le 2x<4), then dividing by (2) gives (-5\le x<2). Check equality at each endpoint separately.

Inequalities in opposite directions do not give a fixed comparison between (a+c) and (b+d). Same direction is the safe rule for adding inequalities.

The direct answer is A: the interval is (-1,3). An interval records which real numbers are allowed. The signs -1<x and x<3 are strict inequalities, so x may be greater than -1 and less than 3, but it may not equal either endpoint. Round parentheses show exclusion. Option A is correct because both -1 and 3 are excluded. Option B, [-1,3], includes both endpoints and would represent -1≤x≤3. Option C, (-1,3], incorrectly includes 3. Option D, [-1,3), incorrectly includes -1. Test an interior value such as 0: it belongs; test either endpoint: neither belongs. Memory cue: strict signs < or > use open ends; ≤ or ≥ use closed square ends.

The values (2) and (8) are excluded, so the natural values are from (3) to (7). Pay close attention to strict signs.

Since p is greater than or equal to q and q is strictly greater than r, the chain p ≥ q > r guarantees p > r. Thus p = r and r ≥ p are impossible, while p < r is in the opposite direction. Exam tip: when inequalities have the same direction, a chain containing at least one strict inequality gives a strict final comparison.

If (k>0), the direction stays the same, and if (k=0), equality occurs. Hence (ka\le kb) is always correct.

For positive (x), equality in (x+\frac{1}{x}\ge 2) occurs at (x=1). Remember the equality condition in standard forms.

The inequality (|x-2|<5) means (-5<x-2<5), so (-3<x<7). A less-than distance inequality gives the middle interval.