\frac{7x + 12}{3} = 5x - 2 \implies 7x + 12 = 15x - 6 \implies -8x = -18 \implies x = \frac{9}{4}.

["# Solving Linear Equations: Step-by-Step Guide with \frac{7x + 12}{3} = 5x - 2", "Solving linear equations is one of the foundational skills in algebra — a cornerstone for advanced mathematics and real-world problem-solving. In this article, we’ll walk through the step-by-step solution to the equation:", "$$\n\frac{7x + 12}{3} = 5x - 2\n$$\nWe’ll uncover how to simplify, manipulate, and solve this equation carefully, illustrating key algebraic principles step-by-step — including how to eliminate fractions, combine like terms, and isolate the variable. Whether you're a student, teacher, or self-learner, this guide will strengthen your understanding of linear equations and build confidence in solving similar problems.", "---", "## Step 1: Eliminate the Fraction", "The equation begins with a fraction:\n$$\n\frac{7x + 12}{3} = 5x - 2\n$$\nTo eliminate the denominator, multiply both sides by 3 — the least common denominator:\n$$\n3 \cdot \frac{7x + 12}{3} = 3(5x - 2)\n$$\nSimplifying both sides gives:\n$$\n7x + 12 = 15x - 6\n$$\nWhy this works: Multiplying both sides by 3 preserves the equality while clearing the fraction, making the equation easier to solve.", "---", "## Step 2: Collect Like Terms", "Now we have:\n$$\n7x + 12 = 15x - 6\n$$\nOur goal is to isolate the variable ( x ). Start by moving all terms involving ( x ) to one side and constant terms to the other.", "Subtract ( 7x ) from both sides:\n$$\n12 = 15x - 7x - 6\n\Rightarrow 12 = 8x - 6\n$$\nNext, add 6 to both sides to eliminate the constant on the right:\n$$\n12 + 6 = 8x\n\Rightarrow 18 = 8x\n$$\nWhy this matters: Grouping like terms organizes the equation and eliminates distractions, paving the way to isolate ( x ).", "---", "## Step 3: Solve for ( x )", "We now have:\n$$\n8x = 18\n$$\nDivide both sides by 8 to solve for ( x ):\n$$\nx = \frac{18}{8}\n$$\nSimplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 2:\n$$\nx = \frac{9}{4}\n$$\nPro tip: Always simplify your final answer to its lowest terms to ensure clarity and precision.", "---", "## Final Answer:", "$$\nx = \frac{9}{4}\n$$\nThis solution represents the unique value of ( x ) that satisfies the original equation.", "---", "## Why This Process Works — Key Algebraic Concepts", "- Multiplying durch into both sides: Maintains equation balance while simplifying expression.\n- Collecting like terms: Central to isolating variables and simplifying equations.\n- Inverse operations: Used to isolate ( x ) (subtracting to undo addition, dividing to undo multiplication).", "Understanding each step strengthens problem-solving skills and prepares you for graphing linear functions, systems of equations, and even calculus.", "---", "## Practice Problems to Reinforce Learning", "Try solving these similar equations to build mastery:\n1. ( \frac{2x - 5}{4} = 3x + 1 )\n2. ( \frac{5x + 3}{2} = 2x - 7 )\n3. ( \frac{4x + 1}{3} + 2 = \frac{3x - 4}{2} )", "Each reflects slightly more complex forms — but the same systematic approach applies.", "---", "## Conclusion", "Mastering linear equations is essential for algebra success. By carefully eliminating fractions, combining like terms, and applying inverse operations, expressions like\n$$\n\frac{7x + 12}{3} = 5x - 2\n$$\ntransition from confusing fractions to a clean, solvable equation. Remember to simplify your answer — every step is part of the journey toward accurate, elegant solutions.", "With consistent practice, you’ll develop an intuitive grasp of solving linear equations — a powerful tool in mathematics and daily life. Keep practicing, keep learning.", "---", "Keywords: linear equations, algebra, solving equations, step-by-step, \frac{7x + 12}{3} = 5x - 2, isolating x, fraction elimination, equivalent equations, fraction simplification.\nMeta Description: Learn step-by-step how to solve \frac{7x + 12}{3} = 5x - 2 by eliminating fractions, collecting like terms, and isolating x. Get a simplified explanation with full working and practice problems."]