Solving Linear Equations A Step-by-Step Guide To Carina's Problem

by THE IDEN 66 views

#H1 Introduction

When delving into the realm of mathematics, solving linear equations stands as a fundamental skill. Linear equations, characterized by a single variable raised to the power of one, are ubiquitous in various mathematical and scientific applications. This article will explore the step-by-step process of solving a particular linear equation, providing a comprehensive understanding of the techniques involved. We will dissect the equation Carina is tackling, which is -4- rac{2}{3}x = -6, and shed light on the subsequent steps required to arrive at the solution. Our focus will be on the underlying principles and the rationale behind each operation, ensuring that the reader grasps the mechanics of solving linear equations with confidence. Understanding these concepts will empower you to tackle similar problems and appreciate the elegance of algebraic manipulations. So, let's embark on this mathematical journey and unravel the intricacies of solving for the unknown variable.

#H2 Initial Step: Isolating the Term with the Variable

The initial stage in solving the equation -4- rac{2}{3}x = -6 involves isolating the term containing the variable, which in this case is - rac{2}{3}x. Carina correctly initiates the process by adding 4 to both sides of the equation. This operation maintains the equality because adding the same value to both sides does not alter the balance. The underlying principle here is the addition property of equality, which states that if a=ba = b, then a+c=b+ca + c = b + c for any real number cc. By adding 4 to both sides, the equation transforms as follows:

βˆ’4βˆ’23x+4=βˆ’6+4-4 - \frac{2}{3}x + 4 = -6 + 4

Simplifying both sides, we get:

βˆ’23x=βˆ’2-\frac{2}{3}x = -2

This step effectively eliminates the constant term on the left side, bringing us closer to isolating the variable xx. It's crucial to perform the same operation on both sides to preserve the equation's balance. This foundational step sets the stage for the subsequent operations that will ultimately lead to the solution. The isolation of the variable term is a pivotal moment in the solving process, paving the way for the final steps that will unveil the value of xx. Remember, precision in each step is paramount to ensure the accuracy of the final solution. By meticulously applying the properties of equality, we navigate through the equation with clarity and purpose.

#H3 Further Steps: Multiplying by the Reciprocal

After adding 4 to both sides and obtaining the equation βˆ’23x=βˆ’2-\frac{2}{3}x = -2, the next crucial step involves isolating xx completely. To achieve this, we need to eliminate the coefficient βˆ’23-\frac{2}{3} that is multiplying xx. The most efficient way to do this is by multiplying both sides of the equation by the reciprocal of the coefficient. The reciprocal of βˆ’23-\frac{2}{3} is βˆ’32-\frac{3}{2}. This technique leverages the multiplication property of equality, which asserts that if a=ba = b, then ac=bcac = bc for any real number cc. By multiplying both sides by the reciprocal, we ensure that the coefficient of xx becomes 1, effectively isolating the variable.

Multiplying both sides of the equation βˆ’23x=βˆ’2-\frac{2}{3}x = -2 by βˆ’32-\frac{3}{2} yields:

(βˆ’32)Γ—(βˆ’23x)=βˆ’2Γ—(βˆ’32)\left(-\frac{3}{2}\right) \times \left(-\frac{2}{3}x\right) = -2 \times \left(-\frac{3}{2}\right)

On the left side, the fractions cancel out, leaving us with just xx. On the right side, the product of -2 and -\frac{3}{2} is 3. Therefore, the equation simplifies to:

x=3x = 3

This step is pivotal in solving for xx because it directly addresses the coefficient that is hindering the variable's isolation. By skillfully employing the reciprocal, we transform the equation into its simplest form, revealing the value of xx. This maneuver showcases the power of algebraic manipulation in untangling equations and extracting the desired solution. Remember, the goal is always to isolate the variable, and multiplying by the reciprocal is a potent tool in achieving this objective. The precision in applying this step is key to arriving at the correct answer, underscoring the importance of understanding the properties of equality.

#H4 Verifying the Solution

Once we have obtained a solution, it is imperative to verify its correctness. This crucial step ensures that the value we have found for xx indeed satisfies the original equation. Verification involves substituting the solution back into the original equation and checking if both sides of the equation are equal. This process not only validates our solution but also helps identify any potential errors made during the solving process. For the equation βˆ’4βˆ’23x=βˆ’6-4 - \frac{2}{3}x = -6, we found that x=3x = 3. To verify this, we substitute x=3x = 3 back into the original equation:

βˆ’4βˆ’23(3)=βˆ’6-4 - \frac{2}{3}(3) = -6

Simplifying the left side, we get:

βˆ’4βˆ’2=βˆ’6-4 - 2 = -6

βˆ’6=βˆ’6-6 = -6

Since both sides of the equation are equal, the solution x=3x = 3 is verified as correct. This step solidifies our confidence in the solution and confirms that we have accurately navigated through the algebraic manipulations. Verification is a cornerstone of problem-solving in mathematics, providing a safeguard against errors and a deeper understanding of the equation's behavior. It's a testament to the rigor of mathematical processes, where every step is subject to scrutiny and validation. By making verification a standard practice, we enhance our problem-solving skills and cultivate a meticulous approach to mathematics.

#H5 Possible True Statements Regarding the Solving Process

Now, let's analyze the possible true statements regarding the rest of the solving process after Carina adds 4 to both sides of the equation. The equation at this stage is βˆ’23x=βˆ’2-\frac{2}{3}x = -2. We need to identify which statements accurately describe the subsequent steps and the properties involved.

  1. Multiplying both sides by βˆ’32-\frac{3}{2}: This is a valid step, as we discussed earlier. Multiplying by the reciprocal isolates xx. This statement is true because multiplying both sides by βˆ’32-\frac{3}{2} will indeed lead to the solution.
  2. Dividing both sides by βˆ’23-\frac{2}{3}: Dividing by a fraction is equivalent to multiplying by its reciprocal. Therefore, dividing both sides by βˆ’23-\frac{2}{3} is another valid approach to isolate xx. This statement aligns with the mathematical principles and is thus true.
  3. The solution is x=3x = 3: As we verified earlier, the solution to the equation is indeed x=3x = 3. This statement is a direct consequence of the correct application of algebraic manipulations and is therefore true.

These statements encapsulate the key steps and the final solution, providing a comprehensive overview of the solving process. They highlight the importance of understanding the properties of equality and the strategic use of reciprocals in isolating variables. By carefully considering these statements, we reinforce our understanding of linear equation solving and appreciate the logical flow of mathematical reasoning.

#H6 Common Mistakes to Avoid

While solving linear equations, several common mistakes can lead to incorrect solutions. Being aware of these pitfalls can significantly improve accuracy and prevent errors. One frequent mistake is failing to apply the same operation to both sides of the equation. Remember, maintaining balance is crucial; any operation performed on one side must be mirrored on the other. Another common error is mishandling negative signs. Pay close attention to the signs of the terms and coefficients, as a single sign error can propagate through the entire solution process. Additionally, mistakes often occur when dealing with fractions. Ensure that you correctly find reciprocals and perform multiplication or division with fractions. It's also essential to simplify the equation at each step, combining like terms and reducing fractions to their simplest form. This not only makes the equation easier to manage but also reduces the chances of errors. Another oversight is neglecting to verify the solution. Always substitute the obtained value back into the original equation to confirm its correctness. This step serves as a final check and can catch any mistakes made along the way. By being mindful of these common pitfalls and adopting a meticulous approach, you can enhance your problem-solving skills and confidently tackle linear equations.

#H7 Conclusion

In conclusion, solving the linear equation βˆ’4βˆ’23x=βˆ’6-4 - \frac{2}{3}x = -6 involves a series of logical steps rooted in the properties of equality. Carina's initial step of adding 4 to both sides sets the stage for isolating the variable xx. The subsequent steps, such as multiplying by the reciprocal or dividing by the coefficient, are crucial in unveiling the solution. The solution, x=3x = 3, is verified by substituting it back into the original equation, ensuring accuracy. Throughout the process, it's vital to avoid common mistakes such as failing to maintain balance, mishandling negative signs, or incorrectly manipulating fractions. By understanding the underlying principles and adopting a meticulous approach, solving linear equations becomes a manageable and rewarding task. This step-by-step guide provides a solid foundation for tackling more complex mathematical problems and underscores the importance of precision and logical reasoning in mathematics. Remember, practice and attention to detail are key to mastering the art of solving linear equations.