Introduction to inequalities

For every real (x), (x^2\geq 0), so (x^2<0) is impossible. In such questions, check the basic property first.

Since (3) is positive, the sign does not change and (x-2\geq 12) gives (x\geq 14). With positive multiplication, the sign remains the same.

Multiplying by ( -4 ) reverses the sign to (x+1>-8). Be extra careful when the denominator is negative.

(3x-7\leq 2) gives (x\leq 3), and (x=0) satisfies it but not (x<0). Direct option checking is a fast method.

Adding inequalities in the same direction gives (p+r>q+s), which is always true. Subtraction and multiplication need extra conditions.

(5-2x\leq x+11) gives (-3x\leq 6), so (x\geq -2). Division by a negative coefficient reverses the sign.

( |x|<a ) means (-a<x<a) when (a>0). For less-than distance questions, the solution is the middle interval.

( |x-3|\geq 2 ) means (x-3\leq -2) or (x-3\geq 2). Greater-distance questions give two outer intervals.

Multiplying by positive (10) gives (4x-2<5x+20), so (x>-22). With a positive common denominator, the sign does not change.

(2x<11) gives (x<\frac{11}{2}), so the greatest natural number is (x=5). For integer domains, do not write a decimal answer.

In ( (-\infty,4) ), (4) is not included and all smaller values are included. An open dot means the boundary is excluded.

Multiplication or division by a negative number reverses the inequality sign. Addition and subtraction do not change the sign.

Subtracting (1) throughout gives (6\leq 2x<14), then dividing by (2) gives (3\leq x<7). Work on both bounds together.

(-4\leq 2x<10) gives (-2\leq x<5), so (-2,-1,0,1,2,3,4) are (7) solutions. Check endpoints separately.

Since (m) is negative, division by (m) reverses the sign and gives (x<1). In parameter questions, check the sign of the coefficient first.

Both sides are the same (x+3), so the statement becomes (x+3>x+3), which is false. Recognize equal sides quickly.

Since (x^2\geq 0), (x^2+1\geq 1>0) for all real (x). Recognizing always positive forms is useful.

The first inequality gives (x<6), and the second gives (x>-3), so (-3<x<6). In a chained inequality, both conditions must hold together.

More than (40) means (m>40), so (40) is not included. In word problems, identify whether the boundary is included.

Not more than means at most (500), so (p\leq 500). Words like at most and at least are important.

Direct answer: option A, x+y>3. We are told x>1 and y>2. Add the inequalities term by term: x+y>1+2, so x+y>3. This conclusion is always true, although the sum need not equal 3; for example, x=2 and y=3 gives 5. Option A is correct because both quantities are strictly greater than their lower bounds. Option B, x+y<3, is wrong because choosing x=2,y=3 gives a sum of 5. Option C, x+y=3, is wrong because the values can be larger than 1 and 2, and strict inequalities prevent equality with the lower-bound sum. Option D, x+y≤3, is also wrong for the same example and is weaker in the wrong direction. Adding inequalities with the same direction is valid here. Memory cue: greater than plus greater than gives greater than the sum of the bounds.

The direct answer is B: x=−2 is possible. The condition −2<x≤4 has two parts: x must be strictly greater than −2, and x may be equal to 4. Thus −2 is excluded, while 4 is included. A is true because 4≤4. B is false because −2 is not greater than −2; equality is forbidden by the strict < sign. C is true because −2<0≤4. D is true because −2<3.5≤4. On a number line, place an open circle at −2 and a closed circle at 4, shading only the interval between them. The open endpoint represents exclusion, and the closed endpoint represents inclusion. Therefore the statement asking whether x=−2 is possible is the false one. A common mistake is to treat < and ≤ alike; the small extra line in ≤ includes the endpoint. Memory cue: open circle means not included, closed circle means included.

(x^2+4) is always at least (4), so (x^2+4\leq 0) is impossible. Use the minimum value to judge solutions.

The same-direction chain gives (a>b>c), hence (a>c). This is the transitive property of inequalities.

Multiplying by positive (2) gives (-2\leq x+3<10), then subtracting (3) gives (-5\leq x<7). Positive multiplication does not change direction.