Solving Algebraic Equations And Expressions A Comprehensive Guide

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In the realm of mathematics, algebraic equations and expressions serve as fundamental building blocks for problem-solving and analytical reasoning. This article delves into a series of algebraic problems, providing comprehensive solutions and explanations to enhance understanding. We will explore how to solve for variables, evaluate expressions, and form linear equations. This exploration aims to equip readers with the skills to tackle various algebraic challenges, fostering a deeper appreciation for the elegance and utility of algebraic principles. Through detailed explanations and step-by-step solutions, we will unravel the intricacies of these problems, making algebra more accessible and engaging for learners of all levels. Grasping these concepts is crucial not only for academic pursuits but also for real-world applications where logical thinking and problem-solving are paramount.

Solving for Variables

The cornerstone of algebra lies in the ability to solve for unknown variables within equations. This involves manipulating equations to isolate the variable on one side, thereby determining its value. Let's consider the problem: If y/5 = 1, then find the value of y. The fundamental principle here is to perform the same operation on both sides of the equation to maintain equality. To isolate 'y', we multiply both sides of the equation by 5. This yields (y/5) * 5 = 1 * 5, which simplifies to y = 5. This straightforward example illustrates the core concept of solving for a variable: using inverse operations to undo the operations affecting the variable. Understanding this process is crucial for tackling more complex equations. In essence, we are reversing the steps that were initially applied to the variable, ensuring that we maintain the balance of the equation throughout the process. This method is applicable across a wide range of algebraic problems, making it a vital skill for any mathematics student.

Detailed Steps for Solving y/5 = 1

  1. Identify the operation affecting 'y': In this case, 'y' is being divided by 5.
  2. Determine the inverse operation: The inverse of division is multiplication.
  3. Apply the inverse operation to both sides of the equation: Multiply both sides by 5.
  4. (y/5) * 5 = 1 * 5
  5. Simplify the equation: The 5 in the numerator and the 5 in the denominator on the left side cancel each other out, leaving 'y'. On the right side, 1 multiplied by 5 equals 5.
  6. y = 5

Therefore, the value of y is 5. This methodical approach ensures accuracy and clarity in solving algebraic equations. The beauty of algebra lies in its structured approach to problem-solving, where each step logically follows from the previous one, leading to a definitive solution. By mastering these fundamental techniques, learners can confidently tackle increasingly complex mathematical challenges.

Evaluating Expressions

Another critical aspect of algebra is evaluating expressions. This involves substituting given values for variables and simplifying the expression to obtain a numerical result. Let's explore two examples:

(i) 7(4/7 + 2/7)

This expression involves a combination of arithmetic operations and fractions. The first step is to simplify the expression within the parentheses. Since the fractions 4/7 and 2/7 have a common denominator, we can add them directly: 4/7 + 2/7 = 6/7. Now the expression becomes 7(6/7). To evaluate this, we multiply 7 by 6/7. This can be visualized as (7/1) * (6/7). The 7 in the numerator and the 7 in the denominator cancel each other out, leaving us with 6. Thus, the value of the expression 7(4/7 + 2/7) is 6. This example highlights the importance of following the order of operations and simplifying fractions to arrive at the correct answer. Understanding how to manipulate fractions and perform arithmetic operations within algebraic expressions is crucial for more advanced mathematical concepts.

Detailed Breakdown of 7(4/7 + 2/7)

  1. Simplify the expression within the parentheses:
    • Find a common denominator for the fractions (in this case, it's already 7).
    • Add the numerators: 4 + 2 = 6.
    • The expression within the parentheses simplifies to 6/7.
  2. Multiply the result by 7:
    • 7 * (6/7)
    • Visualize this as (7/1) * (6/7).
  3. Cancel out common factors:
    • The 7 in the numerator and the 7 in the denominator cancel each other out.
  4. Simplify:
    • This leaves us with 6.

(ii) 7x(1/2 + 3/7)

This expression introduces a variable 'x', which means the final result will be an algebraic term rather than a numerical value. We follow a similar approach as before, first simplifying the expression within the parentheses. To add 1/2 and 3/7, we need a common denominator. The least common multiple of 2 and 7 is 14. So, we convert 1/2 to 7/14 and 3/7 to 6/14. Adding these gives us 7/14 + 6/14 = 13/14. Now the expression becomes 7x(13/14). To simplify this, we multiply 7x by 13/14. This can be written as (7x/1) * (13/14). We can simplify by dividing both 7 and 14 by 7, which gives us (x/1) * (13/2). Multiplying these fractions results in 13x/2. Therefore, the value of the expression 7x(1/2 + 3/7) is 13x/2. This example demonstrates how to handle variables within expressions and the importance of finding common denominators when adding fractions. Mastering these techniques is essential for simplifying and solving more complex algebraic problems.

Step-by-Step Simplification of 7x(1/2 + 3/7)

  1. Simplify the expression within the parentheses:
    • Find the least common denominator (LCD) of 2 and 7, which is 14.
    • Convert 1/2 to an equivalent fraction with a denominator of 14: (1/2) * (7/7) = 7/14.
    • Convert 3/7 to an equivalent fraction with a denominator of 14: (3/7) * (2/2) = 6/14.
    • Add the fractions: 7/14 + 6/14 = 13/14.
  2. Multiply the result by 7x:
    • 7x * (13/14)
    • Visualize this as (7x/1) * (13/14).
  3. Cancel out common factors:
    • The 7 in the numerator and the 14 in the denominator can be simplified by dividing both by 7, resulting in 1 and 2 respectively.
  4. Simplify:
    • This leaves us with (x/1) * (13/2).
    • Multiply the numerators and the denominators: (x * 13) / (1 * 2) = 13x/2.

Substituting Values into Expressions

Another crucial skill in algebra is substituting given values for variables within an expression and then evaluating the result. This combines the concepts of variable substitution and expression simplification. Consider the problem: If p = 3/16, q = -4, then find (cp + tq) - (p - q). Note: There seems to be a typo with 'cp + tq'. Assuming it should be '3p + 2q', let's proceed with the corrected expression.

Corrected Expression: If p = 3/16, q = -4, then find (3p + 2q) - (p - q)

First, we substitute the given values of p and q into the expression. This gives us (3(3/16) + 2(-4)) - (3/16 - (-4)). Next, we simplify each part of the expression separately. 3(3/16) is equal to 9/16, and 2(-4) is equal to -8. So, the first part of the expression becomes 9/16 - 8. To combine these, we need a common denominator. We can rewrite -8 as -128/16. Thus, 9/16 - 128/16 = -119/16. Now let's simplify the second part of the expression, (3/16 - (-4)). Subtracting a negative is the same as adding, so this becomes 3/16 + 4. We can rewrite 4 as 64/16. Therefore, 3/16 + 64/16 = 67/16. Finally, we subtract the second simplified part from the first: -119/16 - 67/16. This simplifies to -186/16, which can be further reduced to -93/8. Therefore, the value of the expression (3p + 2q) - (p - q) when p = 3/16 and q = -4 is -93/8. This example demonstrates the importance of careful substitution and step-by-step simplification to arrive at the correct result.

Detailed Solution for (3p + 2q) - (p - q) with p = 3/16, q = -4

  1. Substitute the values of p and q into the expression:
    • (3(3/16) + 2(-4)) - (3/16 - (-4))
  2. Simplify the first set of parentheses:
    • 3(3/16) = 9/16
    • 2(-4) = -8
    • So, the first part becomes (9/16 - 8)
    • Convert -8 to a fraction with a denominator of 16: -8 = -128/16
    • 9/16 - 128/16 = -119/16
  3. Simplify the second set of parentheses:
    • 3/16 - (-4)
    • Subtracting a negative is the same as adding: 3/16 + 4
    • Convert 4 to a fraction with a denominator of 16: 4 = 64/16
    • 3/16 + 64/16 = 67/16
  4. Subtract the second result from the first:
    • -119/16 - 67/16
    • Combine the fractions: (-119 - 67)/16 = -186/16
  5. Simplify the fraction:
    • Divide both the numerator and the denominator by their greatest common divisor, which is 2.
    • -186/16 = -93/8

Forming Linear Equations

The ability to translate word problems into mathematical equations is a crucial skill in algebra. This involves identifying the unknowns, representing them with variables, and expressing the relationships described in the problem as equations. Let's consider the example: (i) 4m = 7m - 2 and (ii) 3m = 5m - 8/3. These are already given as simple linear equations, but the task might be to solve them. However, let's address the prompt to form equations from word problems with an example.

Example Word Problem: Two-thirds of a number exceeds the number by 5. Form an equation to represent this.

To solve this, we first identify the unknown, which is “a number”. Let’s represent this number with the variable 'x'. “Two-thirds of a number” can be written as (2/3)x. The word “exceeds” implies a subtraction and an inequality. The problem states that two-thirds of the number is 5 more than the number itself. This can be translated into the equation (2/3)x = x + 5. To solve this equation, we would typically subtract x from both sides and then multiply by 3 to eliminate the fraction. However, the focus here is on forming the equation, which we have successfully done. Forming equations from word problems requires careful reading and understanding of the relationships described. It is a fundamental skill in algebra and essential for solving real-world problems.

Detailed Steps to Form the Equation

  1. Identify the unknown: The unknown is “a number”.
  2. Represent the unknown with a variable: Let the number be 'x'.
  3. Translate the words into mathematical expressions:
    • “Two-thirds of a number” translates to (2/3)x.
    • “Exceeds the number by 5” implies that (2/3)x is equal to the number plus 5, which can be written as x + 5.
  4. Form the equation:
    • Combine the expressions to form the equation: (2/3)x = x + 5.

Solving the Given Linear Equations

Now, let's solve the linear equations provided in the original prompt:

(i) 4m = 7m - 2

To solve this equation, we want to isolate the variable 'm'. We can start by subtracting 4m from both sides of the equation: 4m - 4m = 7m - 4m - 2, which simplifies to 0 = 3m - 2. Next, we add 2 to both sides: 0 + 2 = 3m - 2 + 2, which simplifies to 2 = 3m. Finally, we divide both sides by 3 to solve for 'm': 2/3 = 3m/3, which gives us m = 2/3. Therefore, the solution to the equation 4m = 7m - 2 is m = 2/3. This demonstrates a common strategy for solving linear equations: rearranging terms to isolate the variable.

Step-by-Step Solution for 4m = 7m - 2

  1. Subtract 4m from both sides:
    • 4m - 4m = 7m - 4m - 2
    • 0 = 3m - 2
  2. Add 2 to both sides:
    • 0 + 2 = 3m - 2 + 2
    • 2 = 3m
  3. Divide both sides by 3:
    • 2/3 = 3m/3
    • m = 2/3

(ii) 3m = 5m - 8/3

Following a similar approach, we first subtract 3m from both sides: 3m - 3m = 5m - 3m - 8/3, which simplifies to 0 = 2m - 8/3. Next, we add 8/3 to both sides: 0 + 8/3 = 2m - 8/3 + 8/3, which simplifies to 8/3 = 2m. Finally, we divide both sides by 2 (or multiply by 1/2) to solve for 'm': (8/3) * (1/2) = 2m * (1/2), which gives us 8/6 = m. This can be further simplified to m = 4/3. Thus, the solution to the equation 3m = 5m - 8/3 is m = 4/3. This example reinforces the importance of maintaining balance in an equation while isolating the variable.

Detailed Solution for 3m = 5m - 8/3

  1. Subtract 3m from both sides:
    • 3m - 3m = 5m - 3m - 8/3
    • 0 = 2m - 8/3
  2. Add 8/3 to both sides:
    • 0 + 8/3 = 2m - 8/3 + 8/3
    • 8/3 = 2m
  3. Divide both sides by 2 (or multiply by 1/2):
    • (8/3) * (1/2) = 2m * (1/2)
    • 8/6 = m
  4. Simplify the fraction:
    • m = 4/3

Conclusion

In this article, we have explored a range of algebraic problems, from solving for variables and evaluating expressions to forming and solving linear equations. Each problem has been dissected step-by-step, emphasizing the underlying principles and techniques. By mastering these fundamental algebraic skills, learners can build a solid foundation for more advanced mathematical concepts and applications. Algebraic proficiency is not only valuable in academic settings but also in numerous real-world scenarios where problem-solving and analytical thinking are essential. Continuous practice and a deep understanding of the core principles will pave the way for success in algebra and beyond.