Let X = (4-7) And Y = (4+7), Then Value Of (x-y/3) + (xy+xy)+(x+y) Is -
Understanding how to evaluate algebraic expressions based on given values of variables is fundamental in mathematics. In this discussion, we will explore the expression involving variables \(X\) and \(Y\), interpret their assigned values, and compute the overall result. This process not only reinforces basic algebraic skills but also demonstrates the importance of careful substitution and order of operations in solving complex expressions.
---
Introduction to the Problem
The problem begins with defining two variables:
- \(X = (4 - 7)\)
- \(Y = (4 + 7)\)
Following these definitions, the goal is to evaluate the expression:
\[
\left( \frac{x - y}{3} \right) + (xy + xy) + (x + y)
\]
and determine its value.
This type of problem is typical in algebra, requiring the following steps:
- Find the numerical values of \(X\) and \(Y\).
- Substitute these values into the expression.
- Simplify step-by-step, respecting the order of operations (PEMDAS/BODMAS).
- Interpret the final answer.
---
Calculating the Values of X and Y
Step 1: Evaluate \(X = (4 - 7)\)
- \(4 - 7 = -3\)
- Therefore, \(X = -3\)
Step 2: Evaluate \(Y = (4 + 7)\)
- \(4 + 7 = 11\)
- Therefore, \(Y = 11\)
\[
\left( \frac{-3 - 11}{3} \right) + (-3 \times 11 + -3 \times 11) + (-3 + 11)
\]
---
Substituting Values into the Expression
The original expression, with the known values, is:
\[
\left( \frac{x - y}{3} \right) + (xy + xy) + (x + y)
\]
becomes
\[
\left( \frac{-3 - 11}{3} \right) + (-3 \times 11 + -3 \times 11) + (-3 + 11)
\]
Notice that in the expression, the variables \(x\) and \(y\) are placeholders, but since we have assigned \(X = -3\) and \(Y = 11\), we will use \(x = -3\), \(y = 11\).
---
Step-by-Step Simplification
Step 1: Simplify \(\frac{x - y}{3}\)
- \(x - y = -3 - 11 = -14\)
- Therefore, \(\frac{-14}{3}\)
Step 2: Simplify \((xy + xy)\)
- First, find \(xy\):
- Since the expression is \(xy + xy\):
Step 3: Simplify \((x + y)\)
- \(x + y = -3 + 11 = 8\)
\[
-\frac{14}{3} + (-66) + 8
\]
---
Final Calculation and Result
Combine the numerical parts
- Convert all terms to a common denominator to combine the fractions and integers:
- Express \(-66\) and \(8\) as fractions with denominator 3:
- Calculate numerator equivalents:
- \(-66 \times 3 = -198\)
- \(8 \times 3 = 24\)
- So, the expression is:
- Combine the numerators:
Conclusion
The value of the entire expression simplifies to:
\[
\boxed{-\frac{188}{3}}
\]
which is approximately \(-62.666...\).
---
Understanding the Significance of the Result
The negative result indicates that, given the initial values and the structure of the expression, the overall value tilts toward the negative side. It highlights the importance of careful substitution and the order of operations in algebra.
Key points to remember:
- Always evaluate the assigned values for variables before substitution.
- Simplify expressions step-by-step, respecting parentheses.
- Convert mixed expressions to common denominators for easier addition/subtraction.
- Negative results are common in algebra, especially when subtracting larger numbers from smaller ones.
---
Practical Applications of Similar Algebraic Evaluations
Understanding how to evaluate such expressions has numerous real-world applications:
- Financial Calculations: Calculating net gains or losses based on cost and revenue variables.
- Engineering: Analyzing forces, voltages, or other parameters where variables are defined with specific values.
- Physics: Computing quantities like energy, velocity, or acceleration based on given parameters.
- Computer Science: Programming algorithms that involve variable substitution and expression evaluation.
---
Summary
To summarize, evaluating the expression:
\[
\left( \frac{x - y}{3} \right) + (xy + xy) + (x + y)
\]
with \(X = -3\) and \(Y = 11\), results in a final value of:
\[
-\frac{188}{3}
\]
which approximates to \(-62.67\).
This process underscores the importance of systematic evaluation, understanding algebraic operations, and careful substitution. Mastery of these skills enables students and professionals alike to solve complex problems efficiently and accurately.
---
Final Thoughts
Algebra remains a foundational pillar in mathematics, with applications spanning numerous fields. By practicing problems like this, learners develop critical thinking, problem-solving skills, and a deeper understanding of variable manipulation. Always approach such problems methodically, verify each step, and ensure clarity in calculations for accurate results.
---
Keywords: algebra, variable substitution, algebraic expression, evaluation, mathematical calculation, step-by-step solution, simplifying expressions, negative numbers, fractions, algebraic expressions with variables