Introduction to inequalities

From (2x+3\le x-4), we get (x\le -7). In exams keep (x)-terms on one side and constants on the other side.

Since (3) is positive, the sign does not change, and (x-1\ge6) gives (x\ge7). In exams first check the sign of the denominator.

From (2x\le8), (x\le4), and with (x\in\mathbb{N}) we get ({1,2,3,4}). In exams always check the domain.

Both conditions are required together, so (2<x<8). In exams and usually means intersection (\cap).

(-3x\le -6), and dividing by (-3) gives (x\ge2). In exams reverse the sign when dividing by a negative coefficient.

Adding the same real number does not change the direction of inequality. In exams the sign of the multiplier or divisor matters.

Multiplying by negative (r) changes (>) into (<). In exams watch for reversal in options containing (r<0).

In (|x|\ge3), the distance from (0) is at least (3). In exams a greater-than absolute value gives two outer parts.

The square of a real number is never negative, so (x^2\ge0) is always true. In exams remember the basic property of squares.

For real (x), (x^2) is never negative. In exams write (\varnothing) for an impossible inequality.

The product is positive when both factors have the same sign. In exams keep the zero points (-3) and (2) open.

The fraction is positive when numerator and denominator have the same sign. In exams (x=-1) makes the denominator zero, so it is excluded.

At (x=3), the denominator becomes (0), so the fraction is undefined. In exams always exclude values that make the denominator zero.

After cancelling (x), we get (2>-5), which is always true. In exams such questions may have all real numbers as the answer.

After cancelling (x), we get (-6\ge1), which is false. In exams a false constant statement means no solution.

Adding (3) to all three parts gives (4<2x\le12), so (2<x\le6). In exams apply each step to all three parts.

Greater than or equal to is written as (m\ge40). In exams when the word equal appears, look for (\ge) or (\le).

The direct answer is A: \(w\le500\). The phrase maximum load of 500 kg means the lift may carry 500 kg or any smaller non-negative load. “Maximum” includes the limiting value, so the correct symbol is \(\le\), read as less than or equal to. Option A correctly allows w=500 as well as smaller values. Option B, \(w<500\), wrongly excludes exactly 500 kg; it would describe a strict limit below 500. Option C, \(w\ge500\), says the load must be at least 500 kg, which is not the meaning of a maximum and would allow heavier loads. Option D, \(w>500\), says the load must exceed the safe limit and is clearly unsafe. Physically, w cannot be negative, but that extra fact is not needed among the options. Memory cue: maximum means “at most,” so use \(\le\); minimum means “at least,” so use \(\ge\).

The direct answer is A: \(3x<21\). Translate each phrase carefully. “A number” is x. “Three times a number” means multiplying x by 3, giving 3x. “Is less than 21” gives the symbol <, so the complete inequality is \(3x<21\). It could also be solved as x<7 by dividing by positive 3, but the question asks for the correct statement and A is the direct translation. Option B, \(3x>21\), reverses “less than” and says greater than. Option C, \(x+3<21\), means 3 is added to x, not that x is multiplied by 3. Option D, \(x/3<21\), means one-third of x is less than 21, which is a different sentence. The order matters: “three times x” is 3x, not x+3 or x/3. Memory cue: times means multiplication; less than means the left expression is smaller.

The direct answer is A: \(x=6\). Solve first: \(2x-3>7\). Add 3 to both sides: \(2x>10\). Divide by positive 2: \(x>5\). Therefore any value greater than 5 works. Test the options directly: for x=6, \(2(6)-3=12-3=9>7\), so A satisfies the inequality. For x=5, the left side is \(10-3=7\), giving 7>7, which is false; the boundary is not included, so B fails. For x=4, the left side is 5, not greater than 7, so C fails. For x=3, the left side is 3, so D fails. The strict sign > is important: x must be strictly greater than 5, not equal to 5. Memory cue: after solving, substitute each offered value into the original inequality, not a changed form.

From (5-4x\ge-7), (x\le3), so (x=4) is not in the solution. In exams notice the phrase does not satisfy.

From (x+3\ge2x-1), (4\ge x), that is (x\le4). In exams understand the direction when rewriting (4\ge x) as (x\le4).

Multiplying all parts by positive (2) gives (-2\le x<8). In exams positive multiplication does not change signs.

Dividing by (-2) reverses both signs and gives (3>x\ge-5), that is (-5\le x<3). In exams rewrite compound inequalities in increasing order.

Multiplying by positive (3) gives (-12<x+2\le6), so (-14<x\le4). In exams preserve open and closed signs separately.