Solving Equations A Comprehensive Guide To Algebraic Solutions

In the realm of mathematics, solving equations is a fundamental skill. It's the cornerstone of algebra and a crucial tool in various fields, from science and engineering to economics and computer science. This comprehensive guide delves into solving various types of equations, providing step-by-step solutions and explanations to enhance your understanding. We will explore linear equations, quadratic equations, radical equations, and exponential equations, equipping you with the knowledge and techniques to tackle a wide range of algebraic problems. Mastering these concepts will not only improve your mathematical abilities but also sharpen your problem-solving skills, which are invaluable in numerous real-world scenarios. So, let's embark on this journey of algebraic exploration and unlock the secrets of equation solving.

21. Solve for x:

1. $2+5-(3x-5)(x+1)=0$

To solve the equation $2+5-(3x-5)(x+1)=0$, our primary goal is to isolate the variable x. This involves simplifying the equation, expanding any products, and rearranging terms to ultimately find the value(s) of x that satisfy the equation. This process often requires a combination of algebraic manipulations, such as distributing, combining like terms, and factoring. The key is to maintain balance by performing the same operation on both sides of the equation, ensuring that the equality remains valid throughout the solution process. Understanding the order of operations (PEMDAS/BODMAS) is crucial to simplifying expressions correctly. By following these principles, we can systematically work towards isolating x and determining the solution(s) to the equation. In this particular case, we will start by simplifying the constant terms and then expanding the product of the binomials. This will lead us to a quadratic equation, which we can solve using various methods such as factoring, completing the square, or the quadratic formula. The final step is to check our solutions by substituting them back into the original equation to ensure they are valid.

Let's break down the steps:

1. Simplify the constant terms: $7 - (3x - 5)(x + 1) = 0$
2. Expand the product: $(3x - 5)(x + 1) = 3x^2 + 3x - 5x - 5 = 3x^2 - 2x - 5$
3. Substitute back into the equation: $7 - (3x^2 - 2x - 5) = 0$
4. Distribute the negative sign: $7 - 3x^2 + 2x + 5 = 0$
5. Combine like terms: $-3x^2 + 2x + 12 = 0$
6. Multiply the equation by -1 to simplify: $3x^2 - 2x - 12 = 0$

Now, we can use the quadratic formula to solve for x: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = -2$, and $c = -12$.

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-12)}}{2(3)} = \frac{2 \pm \sqrt{4 + 144}}{6} = \frac{2 \pm \sqrt{148}}{6} = \frac{2 \pm 2\sqrt{37}}{6} = \frac{1 \pm \sqrt{37}}{3}$

Therefore, the solutions are:

$x = \frac{1 + \sqrt{37}}{3}$ and $x = \frac{1 - \sqrt{37}}{3}$

2. $3x^2 + 5x + 1 = 0$ (Correct to two decimal places.)

In this case, we have a quadratic equation in the standard form $ax^2 + bx + c = 0$. To solve this, we can use the quadratic formula, which is a powerful tool for finding the roots of any quadratic equation. The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Identifying the coefficients a, b, and c correctly is essential for accurate application of the formula. In our equation, $3x^2 + 5x + 1 = 0$, we have $a = 3$, $b = 5$, and $c = 1$. Substituting these values into the quadratic formula allows us to calculate the solutions for x. The discriminant, $b^2 - 4ac$, within the square root, provides valuable information about the nature of the roots. If the discriminant is positive, we have two distinct real roots. If it's zero, we have one real root (a repeated root). And if it's negative, we have two complex roots. Once we obtain the solutions using the formula, we can approximate them to two decimal places as requested in the problem. This involves using a calculator or long division to find the decimal values and rounding them appropriately. Always double-check your calculations to ensure accuracy, especially when dealing with square roots and fractions.

Let's apply the quadratic formula:

$x = \frac{-5 \pm \sqrt{5^2 - 4(3)(1)}}{2(3)} = \frac{-5 \pm \sqrt{25 - 12}}{6} = \frac{-5 \pm \sqrt{13}}{6}$

So, the solutions are:

$x = \frac{-5 + \sqrt{13}}{6} \approx -0.23$ and $x = \frac{-5 - \sqrt{13}}{6} \approx -1.43$

3. $3

Solving radical equations, such as $3\sqrt{x+2}-1=2x-6$, requires a strategic approach to eliminate the radical and isolate the variable x. The first crucial step is to isolate the radical term on one side of the equation. This involves performing algebraic manipulations such as adding or subtracting terms from both sides. Once the radical is isolated, we can eliminate it by raising both sides of the equation to the appropriate power. In this case, since we have a square root, we will square both sides. This step is critical but also introduces a potential pitfall: extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original radical equation. Therefore, it's essential to check all solutions obtained after eliminating the radical by substituting them back into the original equation. If a solution doesn't satisfy the original equation, it's an extraneous solution and must be discarded. After eliminating the radical, we typically end up with a polynomial equation, which can be solved using standard techniques such as factoring or the quadratic formula. Remember to simplify the equation after each step to make the subsequent steps easier. By carefully following these steps and checking for extraneous solutions, we can accurately solve radical equations.

Here's how to solve this equation:

1. Isolate the radical term: $3\sqrt{x + 2} = 2x - 5$
2. Divide by 3: $\sqrt{x + 2} = \frac{2x - 5}{3}$
3. Square both sides: $(\sqrt{x + 2})^2 = (\frac{2x - 5}{3})^2$
4. Simplify: $x + 2 = \frac{4x^2 - 20x + 25}{9}$
5. Multiply by 9: $9(x + 2) = 4x^2 - 20x + 25$
6. Expand and rearrange: $9x + 18 = 4x^2 - 20x + 25 ightarrow 4x^2 - 29x + 7 = 0$

Now, use the quadratic formula:

$x = \frac{29 \pm \sqrt{(-29)^2 - 4(4)(7)}}{2(4)} = \frac{29 \pm \sqrt{841 - 112}}{8} = \frac{29 \pm \sqrt{729}}{8} = \frac{29 \pm 27}{8}$

So, the solutions are:

$x = \frac{29 + 27}{8} = \frac{56}{8} = 7$ and $x = \frac{29 - 27}{8} = \frac{2}{8} = \frac{1}{4}$

Check for extraneous solutions:

• For $x = 7$: $3\sqrt{7 + 2} - 1 = 3\sqrt{9} - 1 = 3(3) - 1 = 8$ and $2(7) - 6 = 14 - 6 = 8$. This solution is valid.
• For $x = \frac{1}{4}$: $3\sqrt{\frac{1}{4} + 2} - 1 = 3\sqrt{\frac{9}{4}} - 1 = 3(\frac{3}{2}) - 1 = \frac{9}{2} - 1 = \frac{7}{2}$ and $2(\frac{1}{4}) - 6 = \frac{1}{2} - 6 = -\frac{11}{2}$. This solution is extraneous.

Therefore, the solution is $x = 7$.

4. $2+4-\frac{x}{x-1}-\frac{2x+4}{3x+1}=-4$

Solving rational equations, such as $2+4-\frac{x}{x-1}-\frac{2x+4}{3x+1}=-4$, requires a systematic approach to eliminate the fractions and obtain a solvable equation. The first crucial step is to identify the least common denominator (LCD) of all the fractions in the equation. The LCD is the smallest expression that is divisible by all the denominators. Once the LCD is found, we multiply both sides of the equation by it. This step effectively clears the fractions, transforming the rational equation into a more manageable polynomial equation. However, a critical consideration when dealing with rational equations is the identification of restricted values. Restricted values are values of the variable that make any of the denominators in the original equation equal to zero. These values are excluded from the solution set because division by zero is undefined. After clearing the fractions and simplifying the equation, we solve the resulting polynomial equation using standard techniques. Finally, it's essential to check the solutions obtained against the restricted values. Any solution that coincides with a restricted value is an extraneous solution and must be discarded. By carefully following these steps and paying close attention to restricted values, we can accurately solve rational equations.

Let's solve this:

1. Simplify the left side: $6-\frac{x}{x-1}-\frac{2x+4}{3x+1}=-4$
2. Add -6 on both sides: $-\frac{x}{x-1}-\frac{2x+4}{3x+1}=-10$
3. Multiply both sides by -1: $\frac{x}{x-1}+\frac{2x+4}{3x+1}=10$
4. Find the LCD: $(x-1)(3x+1)$
5. Multiply both sides by the LCD: $x(3x+1) + (2x+4)(x-1) = 10(x-1)(3x+1)$
6. Expand and simplify: $3x^2 + x + 2x^2 - 2x + 4x - 4 = 10(3x^2 + x - 3x - 1)$
7. Further simplification: $5x^2 + 3x - 4 = 10(3x^2 - 2x - 1)$
8. Expand the right side: $5x^2 + 3x - 4 = 30x^2 - 20x - 10$
9. Move all terms to one side: $0 = 25x^2 - 23x - 6$

Use the quadratic formula:

$x = \frac{23 \pm \sqrt{(-23)^2 - 4(25)(-6)}}{2(25)} = \frac{23 \pm \sqrt{529 + 600}}{50} = \frac{23 \pm \sqrt{1129}}{50}$

So, the solutions are:

$x = \frac{23 + \sqrt{1129}}{50}$ and $x = \frac{23 - \sqrt{1129}}{50}$

5. $7^{2-t}=4$

Solving exponential equations, such as $7^{2-t}=4$, involves isolating the exponential term and then using logarithms to solve for the variable. The fundamental principle behind solving these equations is the inverse relationship between exponential and logarithmic functions. If we have an equation of the form $a^x = b$, where a is the base, x is the exponent, and b is the result, we can take the logarithm of both sides with base a to isolate x. In practice, we often use the common logarithm (base 10) or the natural logarithm (base e) because these logarithms are readily available on calculators. Applying the logarithm property $\log_a(a^x) = x$ allows us to bring the exponent down and solve for the variable. In this specific equation, $7^{2-t}=4$, we can take the logarithm of both sides with base 7, or we can use the common or natural logarithm. After applying the logarithm, we use algebraic manipulations to isolate the variable t and find its value. It's crucial to remember the properties of logarithms and exponents to effectively solve these types of equations. Always check your solution by substituting it back into the original equation to ensure it satisfies the equation.

To solve this, we'll use logarithms:

1. Take the natural logarithm of both sides: $\ln(7^{2-t}) = \ln(4)$
2. Use the power rule of logarithms: $(2-t)\ln(7) = \ln(4)$
3. Divide by $\ln(7)$: $2 - t = \frac{\ln(4)}{\ln(7)}$
4. Isolate t: $t = 2 - \frac{\ln(4)}{\ln(7)}$

Using a calculator, we find:

$t \approx 2 - 0.712 \approx 1.288$

6. $6(-2x-5)(x+1) = 3$

To solve the equation $6(-2x-5)(x+1) = 3$, our goal is to find the value(s) of x that satisfy the equation. This involves simplifying the equation, expanding any products, and rearranging terms to ultimately isolate the variable x. We can start by dividing both sides of the equation by 6 to simplify the equation. Then, we expand the product of the binomials $(-2x-5)$ and $(x+1)$. This expansion will result in a quadratic expression, which we can further simplify by combining like terms. Once we have a simplified quadratic equation, we can solve it using various methods, such as factoring, completing the square, or the quadratic formula. Each of these methods has its own advantages and may be more suitable depending on the specific form of the quadratic equation. Factoring is often the quickest method if the quadratic expression can be easily factored. The quadratic formula is a general method that can be applied to any quadratic equation. Completing the square is a useful technique for rewriting the quadratic equation in a form that makes it easier to solve. After finding the solutions, it's always a good practice to check them by substituting them back into the original equation to ensure they are valid.

Here's a step-by-step solution:

1. Divide both sides by 6: $(-2x - 5)(x + 1) = \frac{1}{2}$
2. Expand the product: $-2x^2 - 2x - 5x - 5 = \frac{1}{2}$
3. Simplify: $-2x^2 - 7x - 5 = \frac{1}{2}$
4. Multiply by 2 to eliminate the fraction: $-4x^2 - 14x - 10 = 1$
5. Move all terms to one side: $0 = 4x^2 + 14x + 11$

Now, use the quadratic formula:

$x = \frac{-14 \pm \sqrt{14^2 - 4(4)(11)}}{2(4)} = \frac{-14 \pm \sqrt{196 - 176}}{8} = \frac{-14 \pm \sqrt{20}}{8} = \frac{-14 \pm 2\sqrt{5}}{8} = \frac{-7 \pm \sqrt{5}}{4}$

Therefore, the solutions are:

$x = \frac{-7 + \sqrt{5}}{4}$ and $x = \frac{-7 - \sqrt{5}}{4}$

22. Solving Systems of Equations: A Comprehensive Guide

Introduction to Systems of Equations

Solving systems of equations is a fundamental concept in mathematics with applications spanning various fields, including science, engineering, economics, and computer science. A system of equations is a set of two or more equations containing the same variables. The solution to a system of equations is the set of values for the variables that satisfy all equations simultaneously. In simpler terms, it's the point where the graphs of the equations intersect. Understanding how to solve systems of equations is crucial for modeling real-world problems and finding optimal solutions. For instance, in economics, systems of equations can be used to model supply and demand curves, and the solution represents the equilibrium point. In engineering, they can be used to analyze circuits or solve structural problems. The ability to solve systems of equations efficiently and accurately is a valuable skill in many disciplines. There are several methods for solving systems of equations, each with its own advantages and disadvantages. The choice of method often depends on the specific system of equations and the desired level of accuracy. We will explore some of these methods in detail, providing step-by-step explanations and examples to help you master this essential mathematical skill.

Methods for Solving Systems of Equations

There are several methods for solving systems of equations, each with its own strengths and weaknesses. The choice of method often depends on the specific system of equations and the desired level of accuracy. Here are some of the most common methods:

• Substitution Method: The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable, resulting in a single equation with one variable, which can be solved using standard techniques. Once the value of one variable is found, it can be substituted back into either of the original equations to find the value of the other variable. The substitution method is particularly useful when one of the equations can be easily solved for one variable in terms of the other. This method is also effective for solving systems of nonlinear equations, where other methods may not be applicable. However, the substitution method can become cumbersome if the equations are complex or if it's difficult to isolate a variable. Careful algebraic manipulation is essential to avoid errors and ensure accurate solutions. The key is to choose the equation and variable that will lead to the simplest expression after substitution.
• Elimination Method (Addition/Subtraction Method): The elimination method, also known as the addition or subtraction method, involves manipulating the equations so that the coefficients of one variable are opposites or the same. Then, by adding or subtracting the equations, one variable is eliminated, resulting in a single equation with one variable. This equation can be solved using standard techniques, and the value obtained can be substituted back into one of the original equations to find the value of the other variable. The elimination method is particularly effective when the coefficients of one variable are easily made opposites or the same by multiplying one or both equations by a constant. This method is often preferred over substitution when the equations are in standard form ($Ax + By = C$) and none of the variables are easily isolated. The key to success with the elimination method is to carefully choose the multiplier(s) and ensure that the signs are correct when adding or subtracting the equations. A systematic approach will minimize the chances of errors and lead to accurate solutions.
• Graphing Method: The graphing method involves plotting the equations on a coordinate plane and finding the point(s) of intersection. The coordinates of the intersection point(s) represent the solution(s) to the system of equations. This method is particularly useful for visualizing the solution and understanding the relationship between the equations. It is also effective for systems of nonlinear equations, where other methods may be difficult to apply. However, the graphing method has limitations in terms of accuracy. If the intersection point is not a clear integer value, it may be difficult to determine the exact solution from the graph. In such cases, the graphing method can provide an approximate solution, which can then be refined using other methods. Graphing can be done manually or using graphing calculators or software, which can improve the accuracy of the solution. Despite its limitations, the graphing method is a valuable tool for gaining insights into the behavior of systems of equations.
• Matrix Method: The matrix method is a powerful technique for solving systems of linear equations, especially those with three or more variables. This method involves representing the system of equations in matrix form and then using matrix operations, such as Gaussian elimination or matrix inversion, to solve for the variables. The matrix method provides a systematic and efficient way to solve complex systems of equations. Gaussian elimination involves transforming the matrix into row-echelon form or reduced row-echelon form, which allows for easy back-substitution to find the solutions. Matrix inversion involves finding the inverse of the coefficient matrix and then multiplying it by the constant matrix to obtain the solutions. The matrix method is particularly useful in computer programming and numerical analysis, where large systems of equations need to be solved efficiently. While it may seem more abstract than other methods, the matrix method provides a powerful tool for handling complex systems of equations.

Each of these methods offers a unique approach to solving systems of equations, and the best method to use depends on the specific problem at hand. Understanding the strengths and weaknesses of each method allows you to choose the most efficient and accurate approach.

By understanding these methods, you can effectively tackle a wide variety of equation-solving challenges.

This comprehensive guide provides a solid foundation for mastering the art of solving equations. Remember, practice is key to success in mathematics. The more you practice, the more confident and proficient you will become in solving equations.