Introduction to inequalities

Simplifying gives (-8x+6\geq 10), then (-8x\geq 4), so (x\leq -\frac{1}{2}). Reverse the sign when dividing by a negative coefficient.

Multiplying by (6) gives (4x-2\leq x+5), so (3x\leq 7). Multiplying by a positive LCM does not change the inequality sign.

From (7-4x>19), we get (-4x>12), so (x<-3). The inequality sign reverses when dividing by a negative number.

Simplifying gives (10x-15\leq 3x+20), so (7x\leq 35). After opening brackets, place like terms correctly.

Adding (6) gives (-3\leq 3x<18), then (-1\leq x<6). Apply the same operation to all parts of a compound inequality.

Taking both conditions together gives (2<x\leq 9). Use an open endpoint for a strict sign and a closed endpoint for equality.

Multiplying by (10) gives (4x+6\geq 5x-5), so (x\leq 11). Removing positive denominators does not change the inequality sign.

The inequality gives (5x\leq 25), so (x\leq 5). Taking positive integers in (\mathbb{N}), the solutions are (1) through (5).

Simplifying gives (8-10x<0), so (-10x<-8). Dividing by a negative number gives (x>\frac{4}{5}).

Let the number be (x); the result is (x+8), and it is greater than (3x). First translate the statement into the correct algebraic inequality.

Multiplying by (3) gives (6\leq x+1<15), hence (5\leq x<14). Removing a positive denominator does not change the signs.

In the strict inequality (x<-5), (-5) is not included and smaller numbers are included. So an open circle and left shading are correct.

In ((-3,4]), (0) is included and (-3) is excluded because of the open endpoint. Inclusion is decided by brackets.

To determine whether an inequality has real solutions, simplify it and check whether the resulting statement can be true. If the variable cancels and a false numerical statement remains, no value of the variable can satisfy the inequality.

In option A, subtracting \(x\) from both sides gives \(2>5\), which is false. Therefore there is no real number satisfying it. For comparison, option B becomes \(2<5\), which is always true, and option C becomes \(2\le5\), also always true. Option D is \(x+2\ge x+2\), an identity true for every real number. Hence option A is the only no-solution inequality.

From (5x-4\leq 5x+2), we get (-4\leq 2), which is always true. If a true constant statement remains, all real numbers are solutions.

“Not less than 40” means the marks can be equal to 40 or greater than 40. Hence, the correct inequality is \(m\geq 40\). Option A incorrectly excludes the case \(m=40\), even though exactly 40 marks satisfy the passing condition. Exam tip: translate “not less than” as \(\geq\) and “not more than” as \(\leq\).

An open circle shows that (4) is not included, and right shading shows greater numbers. Therefore (x>4) is correct.

The direct answer is option C, \(x=-1\). Start with \(\frac{x-2}{3}\le-1\). Since 3 is positive, multiplying both sides by 3 does not reverse the inequality: \(x-2\le-3\). Add 2 to both sides: \(x\le-1\). The question asks for the greatest possible integer, so the largest integer satisfying this condition is \(-1\). Option A, 1, is not allowed because \(1\le-1\) is false. Option B, 0, is also greater than -1 and fails. Option C, -1, satisfies the inequality exactly: \((-1-2)/3=-1\). Option D, -2, is a valid solution but is not the greatest one, because -1 is larger. Since the variable is restricted to integers, choose the endpoint integer. Remember: multiplying or dividing an inequality by a positive number keeps its direction; only a negative number reverses it.

An inequality has every real number as its solution when simplifying it removes the variable and leaves a statement that is always true. If the remaining statement is always false, there are no solutions. This distinction is useful when the same variable expression appears on both sides.

For option C, subtracting \(7x\) from both sides of \(7x-4\leq7x+9\) gives \(-4\leq9\), which is true for every real number. Hence every real \(x\) satisfies it, so the solution set is \((-\infty,\infty)\). The other options reduce to false statements such as \(-1>2\), \(3<1\), or \(6<-8\).

Subtracting (5) gives (-12<2x\leq 8), then (-6<x\leq 4). Apply the same operation to every part of a compound inequality.

Multiplying by (3) gives (-6\leq x-1<9), so (-5\leq x<10). The integers from (-5) to (9) are (15) in total.

Simplifying gives (3x-6<3x-8), that is (-6<-8), which is false. A false constant statement gives solution set (\emptyset).

Multiplying by (6) gives (10-4x\geq x+1), so (9\geq 5x). Therefore (x\leq \frac{9}{5}), so none of the listed options should be correct.

From ( -3x>9 ), we get (x<-3) because division by ( -3 ) reverses the sign. Always check the sign in the last step.

Direct answer: option B, (2,7]. Interval notation records inclusion at each endpoint. The inequality 2<x says x is greater than 2, so 2 is excluded and we use a round, open bracket at the left. The inequality x≤7 says x may equal 7, so 7 is included and we use a square, closed bracket at the right. Together this is (2,7]. Option A, [2,7], incorrectly includes 2. Option C, [2,7), includes 2 and excludes 7, so both endpoint decisions are wrong. Option D, (2,7), correctly excludes 2 but incorrectly excludes 7. On a number line, draw an open circle at 2, a closed circle at 7, and shade between them. The bracket type is determined by the inequality sign.