- Basic Math and Science Algebraic Equations and Inequalities Questions and Answers Flashcards

Solve for x: |2x − 7| = |x + 3|

Answer: x = 10 or x = 4/3

When two absolute values are equal, either the expressions are equal or opposites. Case 1: 2x − 7 = x + 3 → x = 10. Case 2: 2x − 7 = −(x + 3) → 2x − 7 = −x − 3 → 3x = 4 → x = 4/3. Both solutions check out.

Which of the following is the solution set for the inequality (x − 2)(x + 5) < 0?

Answer: −5 < x < 2

The zeros are x = 2 and x = −5. The parabola opens upward, so the product is negative between the roots. Therefore the solution is −5 < x < 2.

If 3x² − 12 = 0 and x > 0, what is the value of 4x³ − 8x?

Answer: 16

From 3x² − 12 = 0, x² = 4, so x = 2 (since x > 0). Then 4x³ − 8x = 4(8) − 8(2) = 32 − 16 = 16.

A system of equations has no solution. Which of the following could be that system?

Answer: 2x + 4y = 6 and x + 2y = 5

For 2x + 4y = 6 and x + 2y = 5: multiplying the second by 2 gives 2x + 4y = 10, which contradicts 2x + 4y = 6. These lines are parallel with no intersection. Option B has the same line (consistent). Option C has one solution. Option D has infinitely many solutions.

If f(x) = x² − 6x + k has exactly one real root, what is the value of k?

Answer: 9

A quadratic has exactly one real root when its discriminant equals zero: b² − 4ac = 0. Here a = 1, b = −6, c = k. So (−6)² − 4(1)(k) = 0 → 36 − 4k = 0 → k = 9.

Solve for x: (x + 1)/(x − 3) ≥ 2, where x ≠ 3

Answer: 3 < x ≤ 7

Rewrite as (x + 1)/(x − 3) − 2 ≥ 0 → (x + 1 − 2(x − 3))/(x − 3) ≥ 0 → (x + 1 − 2x + 6)/(x − 3) ≥ 0 → (7 − x)/(x − 3) ≥ 0. This fraction is ≥ 0 when numerator and denominator have the same sign. 7 − x ≥ 0 when x ≤ 7; x − 3 > 0 when x > 3 (x ≠ 3). Both positive: 3 7 AND x < 3 — impossible. Solution: 3 < x ≤ 7.