Inequalities Practice
Inequalities Practice. Practice questions to deepen understanding of inequalities. Online math practice with full solutions and step-by-step explanations.
Inequalities Practice — solve linear and quadratic inequalities, represent solutions on a number line, and apply sign-flipping rules. Comprehensive practice with detailed explanations.
Solve the inequality: \(x + 3 > 7\)
Subtract 3 from both sides: \(x > 7 - 3\), therefore \(x > 4\)
Solve the inequality: \(2x - 5 \leq 11\)
Add 5 to both sides: \(2x \leq 16\). Divide both sides by 2: \(x \leq 8\)
Solve the inequality: \(-3x > 12\)
Divide by (-3). When dividing or multiplying by a negative number, the inequality sign reverses! Therefore \(x < -4\)
Which representation is correct for \(x \geq 2\) on the number line?
\(\geq\) or \(\leq\) are drawn with a filled circle (including the number). \(x \geq 2\) means from 2 upward, therefore to the right
Which representation is correct for \(x < -1\) on the number line?
\(<\) is drawn with an open circle (not including the number). \(x < -1\) means less than (-1), therefore to the left
Solve the system of inequalities (and): \(x > 2\) and \(x < 5\)
An "and" system requires both conditions to hold simultaneously. x must be greater than 2 and less than 5, therefore \(2 < x < 5\)
Solve the system of inequalities (or): \(x \leq 1\) or \(x \geq 4\)
An "or" system means one condition is sufficient. x can be less than or equal to 1 or greater than or equal to 4
Solve the system: \(x + 2 > 5\) and \(x - 1 < 6\)
From the first inequality: \(x > 3\). From the second inequality: \(x < 7\). Combined: \(3 < x < 7\)
What happens to the inequality sign when both sides are multiplied by (-2)?
When multiplying or dividing both sides of an inequality by a negative number, you must reverse the inequality sign. For example: \(2 < 3\), but \(-4 > -6\)
Example: 2 < 3 → multiply both sides by (-2) → -4 > -6
Solve: \(-2x + 6 < 0\)
Subtract 6 from both sides: \(-2x < -6\). Divide both sides by (-2) and reverse the sign: \(x > 3\)
Solve the quadratic inequality: \((x - 2)(x - 5) > 0\)
The critical points are \(x = 2\) and \(x = 5\). A product of two factors is positive when both are positive (\(x > 5\)) or both are negative (\(x < 2\))
Solve the inequality: \(x^2 - 4 < 0\)
Factor: \((x - 2)(x + 2) < 0\). A product is negative when one factor is positive and the other is negative, that is \(-2 < x < 2\)
Solve: \(x^2 - 9 \geq 0\)
Factor: \((x - 3)(x + 3) \geq 0\). The product is non-negative when both factors have the same sign: both positive (\(x \geq 3\)) or both negative (\(x \leq -3\))
Which system describes the domain: "x is greater than 1 and less than or equal to 6"?
"Greater than 1" is \(x > 1\) (strict). "Less than or equal to 6" is \(x \leq 6\) (with equality). Both together
Solve: \(x^2 + 5x + 6 > 0\)
Factor: \((x + 2)(x + 3) > 0\). Critical points: \(x = -2, x = -3\). The product is positive when both factors have the same sign: \(x < -3\) or \(x > -2\)
Solve the system: \(2x - 3 > 5\) or \(x + 4 < 2\)
From the first inequality: \(2x > 8\), therefore \(x > 4\). From the second inequality: \(x < -2\). An "or" system gives: \(x > 4\) or \(x < -2\)
Solve: \(-x^2 + 4x - 3 > 0\)
Multiply by (-1) and reverse the sign: \(x^2 - 4x + 3 < 0\). Factor: \((x - 1)(x - 3) < 0\). The product is negative when: \(1 < x < 3\)
What is the solution of \(x^2 \leq 0\)?
\(x^2\) is always non-negative. The only way \(x^2 \leq 0\) is if \(x^2 = 0\), that is \(x = 0\)
Solve the system: \(x - 2 > 0\) and \(3x + 6 \leq 12\)
From the first inequality: \(x > 2\). From the second inequality: \(3x \leq 6\), therefore \(x \leq 2\). No number can be both greater than 2 and less than or equal to 2, therefore no solution
⚠️ No solution — contradiction!
Solve: \((x - 1)(x + 2)(x - 3) < 0\)
Critical points: \(x = -2, x = 1, x = 3\). Check the signs in each interval. The product is negative in the intervals: \(x < -2\) or \(1 < x < 3\)