Introduction to inequalities

(x<0) includes all negative real numbers and does not include (0). Therefore ((-\infty,0)) is correct.

The cost inequality is (150+80m\leq 550), which gives (m\leq 5). Remember that not exceed means (\leq).

After multiplying by positive (3), we get (-6<5-x\leq 12), then the direction changes because of the negative part. The correct solution is (-7\leq x<11).

The governing method is to solve the compound inequality before counting the integers. Divide every part of −10 < 2x ≤ 6 by 2. Since 2 is positive, neither inequality sign changes, so −5 < x ≤ 3. The integer values satisfying this interval are −4, −3, −2, −1, 0, 1, 2 and 3. The lower endpoint −5 is excluded because the first sign is strict, whereas the upper endpoint 3 is included because the second sign permits equality. Counting the listed values gives 8 integer solutions, so option B is correct. A undercounts by one, while C and D include values that lie outside the permitted interval. The equivalent number-line picture has an open point at −5 and a closed point at 3.

The relevant principle is the multiplication property of inequalities. Multiplying both sides of a ≤ b by a positive number keeps the inequality direction unchanged. Since k > 0, multiplying by k gives ka ≤ kb, so option B is correct. Option A would be appropriate only if multiplication by a negative number caused a reversal, but k is explicitly positive. Option C is too strong: a and b need not be equal, so their products need not be equal. Adding k to both sides would also preserve direction and yield a + k ≤ b + k, not the reversed statement in option D. Therefore the only conclusion guaranteed for every allowed a, b, and positive k is B.

The word “and” means that both conditions must hold simultaneously, so we take their intersection. The condition x≥−1 includes −1 and all numbers to its right; its interval is [−1,∞). The condition x<4 excludes 4 and includes every number smaller than 4; its interval is (−∞,4). Their common part is [−1,4). The square bracket at −1 shows that −1 is included because the first inequality uses ≥. The parenthesis at 4 shows that 4 is excluded because the second inequality uses <. Therefore option A is correct. Option B reverses the endpoint inclusions, C represents outside values, and D wrongly excludes −1.

The governing sign rule is that a reciprocal has the same sign as its nonzero denominator because the numerator 1 is positive. If x>0, then 1/x>0. If x<0, then dividing the positive number 1 by a negative number gives a negative result, so the inequality fails. The value x=0 is excluded because division by zero is undefined. Therefore the complete solution is x>0, and option A is correct. For instance, x=2 gives 1/2>0, whereas x=−2 gives −1/2<0. Option B has the wrong sign, option C incorrectly includes negative values, and D is false because every positive real x is a valid solution.

Both conditions must be checked separately for each ordered pair. In option B, x=4 and y=8, so x+y=4+8=12>10, satisfying the first inequality. However, x>5 is false because 4 is not greater than 5. Thus (4,8) satisfies the required combination and option B is correct. Option A has sum 11>10 but x=6 does satisfy x>5, so it fails the second requirement. Option C has x=7>5, even though its sum is only 9, and therefore fails both requirements. Option D has sum 10, not greater than 10, and x=5 is not greater than 5. Careful attention to strict signs is essential.

The governing order rule is that division by a positive number preserves the direction of an inequality, whereas division by a negative number reverses it. Since a < b, dividing both sides by c gives a/c < b/c only when c is positive. If c < 0, the correct result would be a/c > b/c. If c = 0, the fractions are undefined, so c = 0 is not allowed. Therefore option A, c > 0, is the only correct condition. Option D is also invalid because it includes negative values, which reverse the inequality, and zero, where division is impossible.

The statement x ≥ 5 means that x is either equal to 5 or greater than 5. Its negation must include every real number that fails this condition and exclude every number that satisfies it. Those values are precisely the numbers less than 5, so the negation is x < 5 and option A is correct. The endpoint is important: x = 5 satisfies the original statement, so it cannot belong to its negation. Option B is too broad because it includes 5, option C is only a subset of the original statement rather than its opposite, and option D describes just one value, not all values that contradict x ≥ 5.

Adding or subtracting the same real number from both sides preserves an inequality, and multiplying both sides by a positive number also preserves its direction. Therefore A and D are always true, and C is always true because 3 is positive. However, multiplying a < b by −1 reverses the order, giving −a > −b, not −a < −b. Thus statement B is not always true and is the correct answer. For example, with a = 1 and b = 2, the premise is true, but −1 < −2 is false. The common error is forgetting that negative multiplication changes the direction.

The negation of an inequality must describe exactly the values that do not satisfy the original statement. The statement x ≤ −3 includes every value less than or equal to −3. Its complement on the real number line consists of values strictly greater than −3, so the negation is x > −3, option A. Notice that the equality case is important: x = −3 belongs to the original statement and therefore cannot belong to its negation. Option B incorrectly includes the boundary value, while option C describes only part of the original set rather than its complement. Option D is only one value and does not represent all excluded values. Thus A is the precise logical negation.

The governing concept is the relationship between a square and absolute value. For every real number x, x² < 9 is equivalent to |x| < 3, because 9 = 3². An absolute-value inequality |x| < a describes all numbers whose distance from zero is less than a; therefore it becomes −a < x < a. Substituting a = 3 gives −3 < x < 3. The endpoints are excluded because x = 3 and x = −3 both give x² = 9, not a value strictly less than 9. Option B describes numbers outside the interval, and options C and D incorrectly include the boundary values or reverse the region.

The governing rule is that multiplying both sides of an inequality by a negative number reverses the inequality sign. From a > b and c < 0, multiply both sides by c. Because c is negative, the direction changes, giving ac < bc. The condition a > b > 0 confirms that a and b are positive, but the reversal specifically depends on c being negative. Equality is impossible: if ac = bc, then c(a − b) = 0; since c is nonzero and a − b is positive, this cannot happen. Option B forgets the sign reversal. Option D is also wrong because the exact relation is strict, not merely non-strict. Hence option A is correct.

An inequality has an empty solution set when it reduces to a contradiction that is false for every possible value of x. In option A, subtracting x from both sides gives 2 < 1. This is always false, so no real number can satisfy x + 2 < x + 1; its solution set is empty. The other options simplify to identities that are always true. Option B becomes 2 > 1, option C becomes 0 < 1, and option D becomes −3 < 0. Each of those statements is true independently of x, so each corresponding solution set is all real numbers. Therefore option A is the only inconsistent inequality and the only one with no solution.

For every real number x, the square x^2 is non-negative, so x^2≥0 always. To satisfy x^2≤0 at the same time, the square must be exactly zero. The only real number whose square is zero is x=0. Hence the real solution set is {0}, making option A correct. The empty set in option B is wrong because x=0 works; the set of all real numbers in option C is wrong because, for example, 1^2=1 is not less than or equal to zero. Option D includes many negative values whose squares are positive, so it also fails.

Let the number be \(x\). Four times the number is \(4x\), and “less than” is represented by \(<\). Therefore, the correct inequality is \(4x<28\). Option B reverses the meaning to “greater than 28,” while C and D represent adding or subtracting 4 from the number. Exam tip: translate “times” as multiplication and “less than” as \(<\).