Math Questions with Answers (Algebra, Absolute Value & Complex Numbers)
Below is a comprehensive set of 20 math questions covering algebra, systems of equations, absolute value equations and inequalities, rational equations, and complex numbers. Detailed, step-by-step solutions are provided directly beneath each question.
Table of Contents
Practice Questions
Question 1 — Mixture Problem
How much water should be added to 20 ounces of a 20% acid solution to reduce it to a 10% solution?
View Step-by-Step Solution
Let \( x \) be the amount of water added in ounces. The amount of pure acid in the original solution is:
\[ 0.20 \times 20 = 4 \text{ ounces} \]The total volume of the new solution will be \( 20 + x \). We want the new concentration to be 10%:
\[ \frac{4}{20 + x} = 0.10 \] \[ 4 = 0.10(20 + x) = 2 + 0.10x \] \[ 2 = 0.10x \implies x = 20 \]Answer: \( \boxed{20 \text{ ounces}} \)
Verification: \( 4/(20 + 20) = 4/40 = 0.10 = 10\% \) ✓
Question 2 — Literal Equation
Solve for \( C \) in the equation: \( B = 2BD + 2CE + 2CL \)
View Step-by-Step Solution
Isolate the terms containing \( C \) on one side:
\[ B - 2BD = 2CE + 2CL \]Factor both sides:
\[ B(1 - 2D) = C(2E + 2L) \]Divide both sides by \( (2E + 2L) \):
\[ C = \frac{B(1 - 2D)}{2E + 2L} \]Answer: \( \boxed{C = \dfrac{B(1 - 2D)}{2E + 2L}} \)
Question 3 — Work Rate Problem
Thomas can paint a house in 6 days, while Joe can paint the same house in 12 days. How long would it take Thomas and Joe, working together, to paint the house?
View Step-by-Step Solution
Individual rates of work per day:
- Thomas's rate: \( \dfrac{1}{6} \) of the house per day
- Joe's rate: \( \dfrac{1}{12} \) of the house per day
Combined rate:
\[ \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \]Working together, they paint \( 1/4 \) of the house per day, so it takes them 4 days.
Answer: \( \boxed{4 \text{ days}} \)
Question 4 — System of Linear Equations
Solve the system of equations:
\[ \begin{cases} 3x - 4y = 25 \\ -2x + y = -10 \end{cases} \]View Step-by-Step Solution
Solve the second equation for \( y \):
\[ y = 2x - 10 \]Substitute into the first equation:
\[ 3x - 4(2x - 10) = 25 \] \[ 3x - 8x + 40 = 25 \] \[ -5x = -15 \implies x = 3 \]Substitute back to find \( y \):
\[ y = 2(3) - 10 = -4 \]Answer: \( \boxed{(3, -4)} \)
Question 5 — Wind and Ground Speed
An airliner travels 3000 miles with a tailwind in 4 hours and returns against the wind in 4.8 hours. Find the ground speed of the airliner.
View Step-by-Step Solution
Let \( v \) be the ground speed of the airliner and \( w \) be the wind speed. Using distance = speed × time:
\[ v + w = \frac{3000}{4} = 750 \] \[ v - w = \frac{3000}{4.8} = 625 \]Adding the two equations eliminates \( w \):
\[ 2v = 1375 \implies v = 687.5 \]Answer: \( \boxed{v = 687.5 \text{ mph}} \)
Note: The wind speed is \( w = 62.5 \) mph.
Question 6 — Union of Intervals
Write the following as a single interval if possible: \( (-9, 2) \cup (1, 30) \)
View Step-by-Step Solution
The first interval covers all numbers from \( -9 \) to \( 2 \); the second interval covers \( 1 \) to \( 30 \). Since \( 1 \lt 2 \), the intervals overlap, forming a continuous stretch:
\[ (-9, 30) \]Answer: \( \boxed{(-9, 30)} \)
Question 7 — Absolute Value Equation
Solve the equation: \( |5x - 2| - 2 = 6 \)
View Step-by-Step Solution
Isolate the absolute value expression:
\[ |5x - 2| = 8 \]Split into two equations:
\[ 5x - 2 = 8 \implies 5x = 10 \implies x = 2 \] \[ 5x - 2 = -8 \implies 5x = -6 \implies x = -\frac{6}{5} \]Answer: \( \boxed{x = 2, \; x = -\dfrac{6}{5}} \)
Question 8 — Absolute Value Inequality (<)
Solve the inequality: \( |4x + 6| \lt 30 \)
View Step-by-Step Solution
Rewrite as a compound inequality:
\[ -30 \lt 4x + 6 \lt 30 \] \[ -36 \lt 4x \lt 24 \] \[ -9 \lt x \lt 6 \]Answer: \( \boxed{(-9, 6)} \)
Question 9 — Absolute Value Inequality (>)
Solve the inequality: \( |4x + 5| - 5 \gt 6 \)
View Step-by-Step Solution
Isolate the absolute value:
\[ |4x + 5| \gt 11 \]Split into two inequalities:
\[ 4x + 5 \gt 11 \implies 4x \gt 6 \implies x \gt \frac{3}{2} \] \[ 4x + 5 \lt -11 \implies 4x \lt -16 \implies x \lt -4 \]Answer: \( \boxed{(-\infty, -4) \cup \left(\dfrac{3}{2}, +\infty\right)} \)
Question 10 — Square Root of a Square
Solve the inequality: \( \sqrt{(3x - 2)^2} \lt 5 \)
View Step-by-Step Solution
Recall that \( \sqrt{A^2} = |A| \). Rewrite:
\[ |3x - 2| \lt 5 \] \[ -5 \lt 3x - 2 \lt 5 \] \[ -3 \lt 3x \lt 7 \] \[ -1 \lt x \lt \frac{7}{3} \]Answer: \( \boxed{\left(-1, \dfrac{7}{3}\right)} \)
Question 11 — Absolute Value Equation (Identity Form)
Solve the equation: \( |-2x + 5| = 2x - 5 \)
View Step-by-Step Solution
The right side is the negative of the inside of the absolute value. By definition, \( |A| = -A \) if and only if \( A \le 0 \). Here, \( A = -2x + 5 \):
\[ -2x + 5 \le 0 \] \[ -2x \le -5 \implies x \ge \frac{5}{2} \]Answer: \( \boxed{\left[\dfrac{5}{2}, +\infty\right)} \)
Question 12 — Absolute Value Equation with Extraneous Solution
Solve the equation: \( |4x - 2| = 8x - 12 \)
View Step-by-Step Solution
Case 1: \( 4x - 2 = 8x - 12 \)
\[ -4x = -10 \implies x = \frac{5}{2} \]Check: LHS \( = |4(2.5) - 2| = 8 \); RHS \( = 8(2.5) - 12 = 8 \). ✓
Case 2: \( 4x - 2 = -(8x - 12) \)
\[ 4x - 2 = -8x + 12 \implies 12x = 14 \implies x = \frac{7}{6} \]Check: LHS \( = |8/3| = 8/3 \); RHS \( = -8/3 \). Absolute value cannot equal a negative number, so this is extraneous.
Answer: \( \boxed{x = \dfrac{5}{2}} \)
Question 13 — Complex Numbers: Multiplication
Evaluate and write in standard form: \( (4 - 2i)(-3 + 3i) \), where \( i^2 = -1 \)
View Step-by-Step Solution
Expand using FOIL:
\[ (4)(-3) + (4)(3i) + (-2i)(-3) + (-2i)(3i) \] \[ -12 + 12i + 6i - 6i^2 \]Substitute \( i^2 = -1 \):
\[ -12 + 18i - 6(-1) = -12 + 18i + 6 = -6 + 18i \]Answer: \( \boxed{-6 + 18i} \)
Question 14 — Complex Numbers: Division
Evaluate and write in standard form: \( \dfrac{2 - 3i}{2 - 2i} \), where \( i^2 = -1 \)
View Step-by-Step Solution
Multiply numerator and denominator by the conjugate \( (2 + 2i) \):
\[ \frac{(2 - 3i)(2 + 2i)}{(2 - 2i)(2 + 2i)} \]Numerator:
\[ 4 + 4i - 6i - 6i^2 = 4 - 2i + 6 = 10 - 2i \]Denominator: \( 2^2 + 2^2 = 8 \)
\[ \frac{10 - 2i}{8} = \frac{5}{4} - \frac{i}{4} \]Answer: \( \boxed{\dfrac{5}{4} - \dfrac{i}{4}} \)
Question 15 — Complex Numbers: Combined Operations
Evaluate and write in standard form: \( (-3 + 2i)^2 - 3(3 - i)(-2 + 2i) \), where \( i^2 = -1 \)
View Step-by-Step Solution
Part 1 — Square the first term:
\[ (-3 + 2i)^2 = 9 - 12i + 4i^2 = 9 - 12i - 4 = 5 - 12i \]Part 2 — Multiply the two binomials:
\[ (3 - i)(-2 + 2i) = -6 + 6i + 2i - 2i^2 = -6 + 8i + 2 = -4 + 8i \] \[ 3(-4 + 8i) = -12 + 24i \]Subtract:
\[ (5 - 12i) - (-12 + 24i) = 5 - 12i + 12 - 24i = 17 - 36i \]Answer: \( \boxed{17 - 36i} \)
Question 16 — Quadratic with Complex Roots
Solve the equation: \( -(x - 2)^2 - 2 = 7 \)
View Step-by-Step Solution
Isolate the squared term:
\[ -(x - 2)^2 = 9 \implies (x - 2)^2 = -9 \]Take the square root of both sides, using \( i \) for the negative radicand:
\[ x - 2 = \pm\sqrt{-9} = \pm 3i \] \[ x = 2 \pm 3i \]Answer: \( \boxed{x = 2 \pm 3i} \)
Question 17 — Quadratic with Radical Roots
Solve the equation: \( (-x + 3)^2 - 12 = 8 \)
View Step-by-Step Solution
Isolate the squared term:
\[ (-x + 3)^2 = 20 \] \[ -x + 3 = \pm\sqrt{20} = \pm 2\sqrt{5} \] \[ -x = -3 \pm 2\sqrt{5} \] \[ x = 3 \mp 2\sqrt{5} \]Answer: \( \boxed{x = 3 \pm 2\sqrt{5}} \)
Question 18 — Rational Equation
Solve the equation: \( 3 + \dfrac{30}{x^2} = -\dfrac{21}{x} \)
View Step-by-Step Solution
Multiply by \( x^2 \) to clear the fractions:
\[ 3x^2 + 30 = -21x \] \[ 3x^2 + 21x + 30 = 0 \]Divide by 3:
\[ x^2 + 7x + 10 = 0 \] \[ (x + 5)(x + 2) = 0 \] \[ x = -5, \quad x = -2 \]Answer: \( \boxed{x = -2, \; x = -5} \)
Question 19 — Rational Equation with Complex Solutions
Solve the equation: \( -\dfrac{x}{x^2 + 2x - 3} + \dfrac{5}{x + 3} = 1 \)
View Step-by-Step Solution
Factor \( x^2 + 2x - 3 = (x + 3)(x - 1) \):
\[ -\frac{x}{(x + 3)(x - 1)} + \frac{5}{x + 3} = 1 \]Multiply by the LCD \( (x + 3)(x - 1) \):
\[ -x + 5(x - 1) = (x + 3)(x - 1) \] \[ -x + 5x - 5 = x^2 + 2x - 3 \] \[ 4x - 5 = x^2 + 2x - 3 \] \[ x^2 - 2x + 2 = 0 \]Quadratic formula:
\[ x = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \]Answer: \( \boxed{x = 1 \pm i} \)
Question 20 — Sum of a Number and Its Reciprocal
The sum of a number and its reciprocal is \( \dfrac{82}{9} \). Find the number.
View Step-by-Step Solution
Let the number be \( x \):
\[ x + \frac{1}{x} = \frac{82}{9} \]Multiply by \( 9x \):
\[ 9x^2 + 9 = 82x \implies 9x^2 - 82x + 9 = 0 \]Factor:
\[ (9x - 1)(x - 9) = 0 \] \[ x = \frac{1}{9} \quad \text{or} \quad x = 9 \]Answer: \( \boxed{x = 9 \text{ or } x = \dfrac{1}{9}} \)
Quick Answer Key
| # | Answer | # | Answer |
|---|---|---|---|
| 1 | 20 oz | 11 | [5/2, +∞) |
| 2 | \( C = B(1-2D)/(2E+2L) \) | 12 | x = 5/2 |
| 3 | 4 days | 13 | −6 + 18i |
| 4 | (3, −4) | 14 | 5/4 − i/4 |
| 5 | 687.5 mph | 15 | 17 − 36i |
| 6 | (−9, 30) | 16 | 2 ± 3i |
| 7 | x = 2, x = −6/5 | 17 | 3 ± 2√5 |
| 8 | (−9, 6) | 18 | x = −2, −5 |
| 9 | (−∞, −4) ∪ (3/2, +∞) | 19 | 1 ± i |
| 10 | (−1, 7/3) | 20 | x = 9 or 1/9 |
Frequently Asked Questions
How much water should be added to 20 ounces of a 20% acid solution to reduce it to a 10% solution?
Calculate the initial amount of acid (\( 0.20 \times 20 = 4 \) ounces). Set up \( \dfrac{4}{20 + x} = 0.10 \), where \( x \) is the water added. Solving gives \( x = 20 \) ounces of water.
How long would it take Thomas and Joe, working together, to paint a house if Thomas takes 6 days and Joe takes 12 days?
Add their daily rates: \( \dfrac{1}{6} + \dfrac{1}{12} = \dfrac{3}{12} = \dfrac{1}{4} \). Their combined rate is \( 1/4 \) of the house per day, so it takes them 4 days working together.
How do you solve an absolute value equation like \( |4x - 2| = 8x - 12 \)?
Split into two cases: (1) \( 4x - 2 = 8x - 12 \), giving \( x = 5/2 \) (valid); (2) \( 4x - 2 = -(8x - 12) \), giving \( x = 7/6 \), which is extraneous because the right side is negative. So the only solution is \( x = 5/2 \).
How do you divide complex numbers?
Multiply both the numerator and denominator by the complex conjugate of the denominator. For \( (2 - 3i)/(2 - 2i) \), multiply by \( (2 + 2i)/(2 + 2i) \) to obtain \( (10 - 2i)/8 = 5/4 - i/4 \).
What is the solution to \( |-2x + 5| = 2x - 5 \)?
Use the identity \( |A| = -A \), which holds when \( A \le 0 \). With \( A = -2x + 5 \), solve \( -2x + 5 \le 0 \) to get \( x \ge 5/2 \). The solution set is \( [5/2, +\infty) \).
How do you solve the equation \( -\dfrac{x}{x^2+2x-3} + \dfrac{5}{x+3} = 1 \)?
Factor the first denominator as \( (x+3)(x-1) \), multiply by the LCD \( (x+3)(x-1) \), and simplify to \( x^2 - 2x + 2 = 0 \). The quadratic formula gives \( x = 1 \pm i \).