2v_3 - 3(6 + 2v_1) = 4 \implies 2v_3 - 18 - 6v_1 = 4 \implies 2v_3 = 22 + 6v_1 \implies v_3 = 11 + 3v_1.
["# Solving 2v₃ − 3(6 + 2v₁) = 4: A Step-by-Step Guide to Isolate v₃", "In algebraic equation solving, breaking down complex expressions methodically is key to finding clear, accurate solutions. Consider the equation:", "[\n2v_3 - 3(6 + 2v_1) = 4\n]", "Understanding how to manipulate and simplify such equations not only aids in finding exact values but also strengthens foundational algebraic skills. Let’s explore step-by-step how to solve for ( v_3 ), uncovering the relationship between variables and revealing simplified expressions.", "## Step 1: Expand the Expression Inside the Parentheses", "The first step is to distribute the -3 across the terms inside the parentheses:", "[\n2v_3 - 3(6 + 2v_1) = 4\n]", "Apply distribution:", "[\n2v_3 - 18 - 6v_1 = 4\n]", "This expansion removes parentheses and organizes constants and variable terms clearly.", "## Step 2: Isolate the Term Containing v₃", "To solve for ( v_3 ), isolate the term with ( v_3 ) by moving other terms to the opposite side:", "[\n2v_3 = 4 + 18 + 6v_1\n]", "Combine like terms on the right side:", "[\n2v_3 = 22 + 6v_1\n]", "## Step 3: Solve for v₃ by Dividing Both Sides by 2", "Now, divide every term by 2 to isolate ( v_3 ):", "[\nv_3 = \frac{22 + 6v_1}{2}\n]", "Simplify the fraction by dividing each term:", "[\nv_3 = 11 + 3v_1\n]", "## Conclusion: Final Expression and Key Insight", "The solution to the equation ( 2v_3 - 3(6 + 2v_1) = 4 ) is:", "[\n\boxed{v_3 = 11 + 3v_1}\n]", "This result shows that ( v_3 ) is a linear function of ( v_1 ), with a constant offset of 11 and a direct proportionality to ( v_1 ) with coefficient 3. Recognizing such transformations is vital in algebra for modeling relationships, simplifying systems, and preparing for larger-scale problem-solving in fields like engineering, economics, and data analysis.", "Mastering these step-by-step manipulations equips learners not just to solve current problems, but also to approach any linear equation with confidence and clarity."]
