Simplify: \(2x + 12x - 15 = 7\), \(14x = 22\), \(x = \frac{22}{14} = \frac{11}{7}\).

Simplify: \(2x + 12x - 15 = 7\), \(14x = 22\), \(x = \frac{22}{14} = \frac{11}{7}\).

["## How to Solve the Equation (2x + 12x - 15 = 7): Step-by-Step Guide", "Solving linear equations like (2x + 12x - 15 = 7) is a foundational skill in algebra. Mastering these steps not only helps with math homework but strengthens logical reasoning used across STEM fields. In this article, we break down the process clearly and simplify the solution: (14x = 22), leading to (x = \frac{11}{7}).", "### Step 1: Combine Like Terms\nThe equation starts with (2x + 12x - 15 = 7). First, combine the like terms on the left side:\n[\n2x + 12x = 14x\n]\nSo the equation becomes:\n[\n14x - 15 = 7\n]", "### Step 2: Isolate the Variable\nNext, eliminate the constant term on the left by adding 15 to both sides:\n[\n14x - 15 + 15 = 7 + 15\n]\nSimplifying gives:\n[\n14x = 22\n]", "### Step 3: Solve for (x)\nTo isolate (x), divide both sides by 14:\n[\nx = \frac{22}{14}\n]", "Reduce the fraction by dividing numerator and denominator by their greatest common divisor, 2:\n[\nx = \frac{11}{7}\n]", "### Why This Method Works\nThis process relies on the basic algebraic principle of performing the same operation on both sides to maintain equation balance. Combining like terms simplifies expressions, while isolating the variable step-by-step leads directly to the solution.", "### Verification\nTo confirm, substitute (x = \frac{11}{7}) back into the original equation:\n[\n2\left(\frac{11}{7}\right) + 12\left(\frac{11}{7}\right) - 15 = \frac{22}{7} + \frac{132}{7} - 15 = \frac{154}{7} - 15 = 22 - 15 = 7\n]\nThe left side matches the right side, confirming the solution is correct.", "### Final Answer\n[\nx = \frac{11}{7}\n]", "---", "If you’re looking to simplify complex equations like this, remember: consolidate terms, isolate variables carefully, and verify your result. With consistent practice, solving linear equations becomes intuitive—empowering you to tackle higher-level math with confidence."]

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