Introduction to inequalities

The inequality \(x\leq 0\) means that \(x\) must be zero or a negative number. For \(x=-3\), the statement \(-3\leq 0\) is true, so option A is correct. All the other options have positive values of \(x\), so they do not satisfy the inequality. Exam tip: The symbol \(\leq\) includes equality as well as less-than values.

(\geq) means greater than or equal to, so (5) is included. On the number line, use a closed circle at (5).

The condition x<0 means that x is negative. The additive inverse of a negative number is positive, so -x>0. Option B incorrectly keeps the same sign, while option D includes zero even though x=0 is excluded. Exam tip: When an inequality is multiplied by -1, its inequality sign reverses.

The boundary value is found by using equality: (2x+5=x+9), so (x=4). The boundary value decides the open or closed endpoint.

Increasing by (3) means (x+3), and greater than means (>). In word problems, define the variable first.

Subtracting (4) from all parts gives (3\leq x<8). Apply the same operation to every part of a compound inequality.

At least means (\geq), so (27+m\geq 40) is correct. In word problems, first identify the mathematical symbol for the key phrase.

Not more than means (\leq), so (2x+5\leq 19) gives (x\leq 7). In word problems, first convert the sentence into an inequality.

Subtracting the same number from both sides does not change the direction. Negative multiplication and squares need extra care.

The governing concept is solving a one-variable linear inequality and translating its solution set into interval notation. Starting with x+9≥2, subtract 9 from both sides to obtain x≥−7. Since equality is allowed, −7 belongs to the solution set and must be written with a square bracket. Values greater than −7 continue without an upper bound, so the right endpoint is infinity, which always uses a parenthesis. Therefore the interval is [−7,∞), option A. Option B wrongly excludes −7, while C and D describe values less than −7 and reverse the direction of the solution.

The direct answer is option A: x ≥ 3. Solve step by step: 11 − 4x ≤ −1. Subtract 11 from both sides to get −4x ≤ −12. Now divide by −4. Division by a negative number reverses the inequality sign, so x ≥ 3. Option A is correct. Option B, x ≤ 3, forgets to reverse the sign and is therefore incorrect. Option C, x > 3, wrongly excludes x = 3; checking x = 3 gives 11 − 12 = −1, which is allowed because equality is included. Option D, x < 3, has both the direction and the endpoint wrong. Always test the boundary: x = 3 works. Memory cue: multiplying or dividing an inequality by a negative reverses its sign.

Adding (1) to all parts gives (3\leq 3x<15), so (1\leq x<5). Apply the same operation to every part of a compound inequality.

The direct answer is A: \(-6<x\leq4\). Begin with \(-8<2x+4\leq12\). Subtract 4 from every part, preserving the inequality directions: \(-12<2x\leq8\). Divide every part by the positive number 2, so the signs do not reverse: \(-6<x\leq4\). Option A matches exactly. Option B changes the strict lower boundary into \(-6\leq x\), wrongly including \(-6\). Option C has boundaries -4 and 6, which do not result from the calculation. Option D has boundaries -2 and 8, also incorrect. The lower endpoint is open because the original lower sign is strict, while the upper endpoint is closed because the original upper sign is inclusive. If division were by a negative number, signs would reverse, but that does not happen here. Memory cue: subtract first, divide by positive 2, and keep the open/closed signs.

For every real number \(x\), we have \(x^2\geq 0\). Therefore, \(x^2+1\geq 1\), which proves that \(x^2+1>0\) is always true. Options B, C and D can be false for particular values; for example, at \(x=0\), B and D are false, while C is never true for a real number. Exam tip: the square of a real number is never negative.

The symbol \(\geq\) means ‘greater than or equal to’: it allows values greater than the given number and also the number itself. The symbol \(>\) excludes equality.

When both sides of an inequality are multiplied by a negative number, the inequality sign reverses. Thus, multiplying \(p<q\) by \(-3\) gives \(-3p>-3q\). Option B is incorrect because it leaves the sign unchanged. Exam tip: always reverse the inequality sign when multiplying or dividing by a negative number.

Because the inequality includes \(-3 \le x\), the value \(-3\) is allowed. The value 2 is not allowed because the condition is strictly \(x<2\). The values 3 and \(-4\) lie outside the given interval. Exam tip: \(\le\) and \(\ge\) include the endpoint, whereas \(<\) and \(>\) exclude it.

Interval notation shows which real numbers satisfy an inequality. The symbol \\(\\leq\\) means that the endpoint is included as well as all numbers smaller than it. The symbol −∞ is not an actual endpoint; it indicates that the interval continues without limit toward smaller numbers. A square bracket at −4 records inclusion of that value.

For x≤−4, x may equal −4 or be any number less than −4. Thus the interval is (−∞,−4], which is option A. The interval with a round bracket at −4 would represent x<−4 and would wrongly exclude −4. Intervals beginning at −4 and extending rightward represent values greater than or equal to −4, not the given inequality.

(-2) is not included, so use a round bracket, and (5) is included, so use a square bracket. Check both boundaries separately.

The direct answer is option A, 6. A boundary value is obtained where the two sides become equal. Replace the inequality by equality: 3x+7=x+19. Subtract x from both sides: 2x+7=19. Subtract 7: 2x=12. Divide by 2: x=6. Thus the boundary value is 6. Option B, 7, is not produced by the algebra. Option C, 8, is also incorrect; substituting it gives unequal sides, 31 and 27. Option D, 5, likewise gives unequal sides, 22 and 24. For the original inequality, solving gives 3x+7\leq x+19, so 2x\leq12 and x\leq6; the boundary is still 6. Memory cue: for a linear inequality, first use equality to locate the boundary, then test the sign.

In the inequality \(0<x\leq 4\), \(x\) must be greater than 0 and less than or equal to 4. Thus, \(x=4\) is valid because the upper endpoint is included. \(x=0\) is invalid because the lower condition is the strict inequality \(x>0\). Exam tip: the signs \(<\) and \(>\) exclude the endpoint, whereas \(\leq\) and \(\geq\) include it.

(-\frac{x}{3}\geq -4), and removing the negative factor gives (x\leq 12). Identify when the sign must be reversed.

Increasing a number by (8) is (x+8), and less than means (<). Identify the main operation in word problems.

(>) does not include the boundary, but (\geq) includes it. This decides the open or closed circle.

Subtracting (2) from all parts gives (3\leq x<9). Apply the same operation to every part of a compound inequality.