Solving Equations And Inequalities A Step By Step Guide

In the realm of mathematics, equations and inequalities serve as fundamental tools for modeling and solving real-world problems. This comprehensive guide delves into the intricacies of solving various types of equations and inequalities, providing step-by-step explanations and practical examples to enhance your understanding. From rational equations to quadratic equations and compound inequalities, we will explore a range of techniques to master these essential mathematical concepts. In this article, we will dissect and solve several mathematical problems, ranging from rational equations and inequalities to exponential equations and word problems. We aim to provide a clear, step-by-step approach to each problem, ensuring a comprehensive understanding of the underlying principles. Whether you're a student tackling algebra or simply looking to refresh your math skills, this guide is designed to equip you with the knowledge and confidence to solve a wide array of mathematical challenges. Let's embark on this mathematical journey together, unraveling the complexities and celebrating the elegance of problem-solving.

6. 3. 1 Problem: ${ \frac{1}{x+3} = \frac{x-1}{-3} }$

To tackle this rational equation, our primary goal is to eliminate the fractions. We achieve this by multiplying both sides of the equation by the least common denominator (LCD), which in this case is -3(x+3). This strategic maneuver clears the denominators, transforming the equation into a more manageable form. Once the fractions are eliminated, we can simplify the equation by performing algebraic operations such as distribution and combining like terms. This process allows us to isolate the variable and ultimately solve for its value. It's crucial to remember that when dealing with rational equations, we must always check our solutions to ensure they do not result in division by zero, which would render the solution extraneous. By following these steps meticulously, we can confidently navigate the complexities of rational equations and arrive at accurate solutions.

Step 1: Eliminate the Fractions

Multiply both sides by -3(x+3):

${ -3(x+3) \cdot \frac{1}{x+3} = -3(x+3) \cdot \frac{x-1}{-3} }$

This simplifies to:

${ -3 = (x+3)(x-1) }$

Step 2: Expand and Simplify

Expand the right side:

${ -3 = x^2 + 2x - 3 }$

Move all terms to one side to set the equation to zero:

${ 0 = x^2 + 2x }$

Step 3: Factor the Quadratic

Factor out the common factor x:

${ 0 = x(x + 2) }$

Step 4: Solve for x

Set each factor equal to zero:

${ x = 0 \text{ or } x + 2 = 0 }$

This gives us two potential solutions:

${ x = 0 \text{ or } x = -2 }$

Step 5: Check for Extraneous Solutions

We need to verify that these solutions do not make the denominator of the original equation equal to zero.

• For x = 0:

  ${ \frac{1}{0+3} = \frac{0-1}{-3} \implies \frac{1}{3} = \frac{-1}{-3} \implies \frac{1}{3} = \frac{1}{3} }$

  This solution is valid.

• For x = -2:

  ${ \frac{1}{-2+3} = \frac{-2-1}{-3} \implies \frac{1}{1} = \frac{-3}{-3} \implies 1 = 1 }$

  This solution is also valid.

Therefore, the solutions are x = 0 and x = -2.

6. 4 Solving Compound Inequalities

6. 4. 1 Problem: ${ -7 < 2 - 3x \leq 5 }$

Compound inequalities, like the one presented, require a nuanced approach to ensure all conditions are met. Our objective is to isolate the variable, but we must do so while maintaining the integrity of the inequality. This involves performing the same operations on all parts of the inequality to preserve the relationships. We begin by strategically adding or subtracting terms to move constants away from the variable term. Then, we address the coefficient of the variable, dividing or multiplying all parts of the inequality by the appropriate value. It's crucial to remember that multiplying or dividing by a negative number reverses the direction of the inequality signs. Finally, we express the solution in interval notation, which provides a concise way to represent the range of values that satisfy the inequality. By carefully applying these steps, we can confidently solve compound inequalities and accurately represent the solution set.

Step 1: Isolate the Term with x

Subtract 2 from all parts of the inequality:

${ -7 - 2 < 2 - 3x - 2 \leq 5 - 2 }$

This simplifies to:

${ -9 < -3x \leq 3 }$

Step 2: Solve for x

Divide all parts by -3. Remember to reverse the inequality signs since we are dividing by a negative number:

${ \frac{-9}{-3} > \frac{-3x}{-3} \geq \frac{3}{-3} }$

This simplifies to:

${ 3 > x \geq -1 }$

Step 3: Rewrite in Standard Form

It's common to write inequalities with the smaller value on the left:

${ -1 \leq x < 3 }$

Step 4: Express the Solution

The solution in interval notation is [-1, 3).

6. 5 Solving Exponential Equations

6. 5. 1 Problem: ${ 3x^2 + 5x = \frac{1}{81} }$

This equation appears to be a mix of polynomial and exponential forms, which might be a typo. Assuming it is meant to be an exponential equation, let’s consider it as ${ 3^{x^2 + 5x} = \frac{1}{81} }$. To solve exponential equations, the key is to express both sides with the same base. This allows us to equate the exponents and transform the equation into a more manageable form. We begin by identifying the common base, which in this case is 3. Then, we rewrite the equation with both sides expressed as powers of 3. Once we have the same base on both sides, we can equate the exponents, resulting in a polynomial equation. Solving this polynomial equation will give us the values of the variable that satisfy the original exponential equation. It's crucial to remember to check our solutions in the original equation to ensure they are valid. By following these steps, we can effectively solve exponential equations and gain a deeper understanding of exponential relationships.

Step 1: Express Both Sides with the Same Base

Rewrite ${ \frac{1}{81} }$ as a power of 3:

${ \frac{1}{81} = \frac{1}{3^4} = 3^{-4} }$

So the equation becomes:

${ 3^{x^2 + 5x} = 3^{-4} }$

Step 2: Equate the Exponents

Since the bases are the same, we can equate the exponents:

${ x^2 + 5x = -4 }$

Step 3: Solve the Quadratic Equation

Move all terms to one side to set the equation to zero:

${ x^2 + 5x + 4 = 0 }$

Factor the quadratic:

${ (x + 1)(x + 4) = 0 }$

Set each factor equal to zero:

${ x + 1 = 0 \text{ or } x + 4 = 0 }$

This gives us two solutions:

${ x = -1 \text{ or } x = -4 }$

Step 4: Check the Solutions

• For x = -1:

  ${ 3^{(-1)^2 + 5(-1)} = 3^{1 - 5} = 3^{-4} = \frac{1}{81} }$

  This solution is valid.

• For x = -4:

  ${ 3^{(-4)^2 + 5(-4)} = 3^{16 - 20} = 3^{-4} = \frac{1}{81} }$

  This solution is also valid.

Therefore, the solutions are x = -1 and x = -4.

Problem: Relative Motion

This word problem presents a classic scenario of relative motion, where two objects are moving towards each other. Our goal is to determine the point at which they meet, which requires a careful analysis of their speeds and the distance between them. We begin by defining variables to represent the unknowns, such as the time it takes for the planes to meet and the distance each plane travels. Then, we use the fundamental relationship between distance, rate, and time (distance = rate × time) to formulate equations that describe the motion of each plane. Since the planes are moving towards each other, their distances will add up to the total distance between the airports. By setting up and solving these equations, we can determine the time it takes for the planes to meet and their respective distances from their departure points. It's essential to interpret the results in the context of the problem, ensuring that our answers make sense in the real-world scenario. By following this systematic approach, we can effectively solve relative motion problems and gain insights into the dynamics of moving objects.

Problem Statement

Plane A takes off from King Shaka International Airport (KSIA) heading towards Cape Town International Airport (CTIA). At the same time, plane B takes off from CTIA heading towards KSIA. (The problem statement is incomplete, we need additional information such as speeds and the distance between the airports to provide a comprehensive solution. Assuming we have the following details:)

• Distance between KSIA and CTIA: 1300 km
• Speed of plane A: 800 km/h
• Speed of plane B: 500 km/h

Step 1: Define Variables

• Let t be the time (in hours) when the planes meet.
• Let d_A be the distance traveled by plane A.
• Let d_B be the distance traveled by plane B.

Step 2: Formulate Equations

Using the formula distance = speed × time, we have:

${ d_A = 800t }$

${ d_B = 500t }$

Since the planes are traveling towards each other, the sum of their distances equals the total distance between the airports:

${ d_A + d_B = 1300 }$

Step 3: Solve the Equations

Substitute the expressions for d_A and d_B into the equation:

${ 800t + 500t = 1300 }$

Combine like terms:

${ 1300t = 1300 }$

Divide by 1300:

${ t = 1 }$

So the planes meet after 1 hour.

Step 4: Find the Distances

Calculate the distances traveled by each plane:

${ d_A = 800 \cdot 1 = 800 \text{ km} }$

${ d_B = 500 \cdot 1 = 500 \text{ km} }$

Step 5: Interpret the Results

The planes meet after 1 hour. Plane A has traveled 800 km, and plane B has traveled 500 km. This means they meet 800 km from KSIA and 500 km from CTIA.

In this comprehensive guide, we've explored a variety of mathematical problems, from solving rational equations and inequalities to tackling exponential equations and word problems. Each problem was approached with a systematic, step-by-step methodology, emphasizing the importance of understanding the underlying principles. By mastering these techniques, you'll be well-equipped to tackle a wide range of mathematical challenges, enhancing your problem-solving skills and fostering a deeper appreciation for the elegance and power of mathematics. Remember, practice is key to success in mathematics, so continue to apply these concepts and explore new problems to solidify your understanding.