Solve Various Algebraic And Complex Number Problems

Solve various algebraic and complex number problems; including equations, systems of equations, and operations on complex numbers

Solve the following equations and problems: Meaningfully compute, justify each step, and interpret the solutions where relevant.

Paper For Above instruction

Introduction

Mathematics encompasses a broad spectrum of topics, including algebra, systems of equations, and complex number operations. Mastery of these topics requires understanding how to manipulate equations, solve for variables, and perform arithmetic with complex numbers. This paper aims to systematically address a set of diverse mathematical problems that test these skills, providing step-by-step solutions, explanations, and interpretations for each. The problems include linear equations, absolute value equations, systems of linear equations, and operations with complex numbers. Such exercises are fundamental for strengthening mathematical reasoning and algebraic fluency, which are essential for advanced studies and practical applications.

Solution to the problems

1. Solve for n: \( \frac{n}{4} - \frac{5}{6} = \frac{5}{12} \)

First, to eliminate the fractions, multiply the entire equation by the least common denominator (LCD) of 12:

  • LCD of 4, 6, and 12 is 12.

Multiplying both sides by 12:

\(12 \times \left( \frac{n}{4} - \frac{5}{6} \right) = 12 \times \frac{5}{12}\)

\(3n - 10 = 5\)

Next, isolate \( n \):

\(3n = 5 + 10 = 15\)

\(n = \frac{15}{3} = 5\)

Solution: \( n = 5 \)

2. Solve for x: \( \frac{2x}{3} - \frac{4x}{5} = \frac{1}{15} \)

Identify the LCD of the denominators 3, 5, and 15, which is 15. Multiply through by 15:

15 \(\times \left( \frac{2x}{3} - \frac{4x}{5} \right) = 15 \times \frac{1}{15}\)

\(5 \times 2x - 3 \times 4x = 1\)

\(10x - 12x = 1\)

\(-2x = 1\)

\(x = -\frac{1}{2}\)

Solution: \( x = -\frac{1}{2} \)

3. Solve for x: \( \frac{9}{4x} + \frac{1}{3} = \frac{5}{2x} \)

Observe that denominators include \( x \). To clear fractions, multiply through by the least common multiple of the denominators, which contains \( 4x \), \( 3 \), and \( 2x \). The LCD is \( 12x \). Multiplying entire equation by \( 12x \):

\(12x \times \frac{9}{4x} + 12x \times \frac{1}{3} = 12x \times \frac{5}{2x}\)

\(3 \times 9 + 4x \times 4 = 6 \times 5\)

\(27 + 16x = 30\)

Subtract 27 from both sides:

\(16x = 3\)

\(x = \frac{3}{16}\)

Solution: \( x = \frac{3}{16} \)

4. Solve \( | 5x - 7 | = 14 \)

This absolute value equation splits into two cases:

  • Case 1: \( 5x - 7 = 14 \)
  • Case 2: \( 5x - 7 = -14 \)

Solving Case 1:

\(5x = 14 + 7 = 21\)

\(x = \frac{21}{5} = 4.2\)

Solving Case 2:

\(5x = -14 + 7 = -7\)

\(x = -\frac{7}{5} = -1.4\)

Solutions: \( x = \frac{21}{5} \) and \( x = -\frac{7}{5} \)

5. Solve the system: \( 2x + 3y = -x + y = 9 \)

Note: The original system seems to be written ambiguously but is likely:

\( 2x + 3y = 9 \) and \( -x + y = 9 \)

From the second equation: \( y = 9 + x \)

Substitute into first:

\( 2x + 3(9 + x) = 9 \)

\( 2x + 27 + 3x = 9 \)

\( 5x + 27 = 9 \)

\( 5x = 9 - 27 = -18 \)

\( x = -\frac{18}{5} = -3.6 \)

Now, find y:

\( y = 9 + (-\frac{18}{5}) = 9 - \frac{18}{5} = \frac{45}{5} - \frac{18}{5} = \frac{27}{5} = 5.4 \)

Solution: \( x = -\frac{18}{5} \), \( y = \frac{27}{5} \)

6. Solve for \( (x, y, z) \):

- x + 2y + 3z = 0


3x - 2y - 2z = 5


4x + y - 3z = -9

Using substitution or matrix methods, set up the system:

Express the system in matrix form and apply methods such as Gaussian elimination. Here, we'll use substitution:

From the first equation:

\( x = 2y + 3z \)

Substitute into second equation:

\( 3(2y + 3z) - 2y - 2z = 5 \)

\( 6y + 9z - 2y - 2z = 5 \)

\( (6y - 2y) + (9z - 2z) = 5 \)

\( 4y + 7z = 5 \)

Similarly, substitute \( x \) into the third equation:

\( 4(2y + 3z) + y - 3z = -9 \)

\( 8y + 12z + y - 3z = -9 \)

\( (8y + y) + (12z - 3z) = -9 \)

\( 9y + 9z = -9 \)

Divide throughout by 9:

\( y + z = -1 \)

Now, solve this system:

  • From \( y + z = -1 \), \( y = -1 - z \)
  • Plug into \( 4y + 7z = 5 \):

\( 4(-1 - z) + 7z = 5 \)

\( -4 - 4z + 7z = 5 \)

\( -4 + 3z = 5 \)

\( 3z = 9 \Rightarrow z=3 \)

Then, \( y = -1 - 3 = -4 \)

And \( x = 2y + 3z = 2(-4) + 3(3) = -8 + 9 = 1 \)

Answer: \( x=1 \), \( y=-4 \), \( z=3 \)

7. Solve: \( xx + 5 = 0 \)

Assuming the notation meant to write: \( x^2 + 5 = 0 \)

Subtract 5 from both sides:

\( x^2 = -5 \)

Take square roots:

\( x = \pm \sqrt{-5} \)

Recall that \( \sqrt{-5} = \sqrt{5}i \), where \( i^2 = -1 \).

Solutions: \( x = \pm i \sqrt{5} \)

8. Add the complex numbers: \( (5 - 2i) + (1 + 6i) \)

Combine like terms:

Real parts: 5 + 1 = 6

Imaginary parts: -2i + 6i = 4i

Sum: \( 6 + 4i \)

9. Subtract the complex numbers: \( (4 + 3i) - (6 + i) \)

Subtract real parts: 4 - 6 = -2

Subtract imaginary parts: 3i - i = 2i

Difference: \( -2 + 2i \)

10. Multiply the complex numbers: \( (4 + 3i)(6 + i) \)

Use distributive property (FOIL):

\(4 \times 6 + 4 \times i + 3i \times 6 + 3i \times i \)

\(24 + 4i + 18i + 3i^2 \)

Recall \( i^2 = -1 \), so:

\( 24 + 4i + 18i - 3 \)

Sum real parts: 24 - 3 = 21

Sum imaginary parts: 4i + 18i = 22i

Product: \( 21 + 22i \)

Conclusion

Through systematic solution methods, we have addressed a variety of algebraic and complex number problems. Each step involved algebraic manipulations, application of fundamental principles, and interpretations of solutions. These problems exemplify key techniques such as clearing denominators, solving absolute value equations, systems of linear equations, and operations with complex numbers. Mastery of such problems enhances problem-solving skills vital for higher mathematics and scientific applications.

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