Subtract second from third: \( (16a + 4b + c) - (4a + 2b + c) = 15 - 3 \Rightarrow 12a + 2b = 12 \Rightarrow 6a + b = 6 \) (eq2)

["Subtract Second from Third in Linear Expressions: Simplify ( (16a + 4b + c) - (4a + 2b + c) = 15 - 3 ) to Get Equations", "When working with algebraic expressions, especially working with differences of expressions, one common technique is subtracting one expression from another to simplify and form equations. Today, we walk through a key algebraic step involving subtracting a linear expression from a second to yield a clearer linear equation — specifically focusing on solving ( (16a + 4b + c) - (4a + 2b + c) = 15 - 3 ) step-by-step.", "---", "### Understanding the Problem", "We start with the expression equation:\n[\n(16a + 4b + c) - (4a + 2b + c) = 15 - 3\n]", "This means we compute the difference between two linear expressions on the left-hand side and simplify the constant values on the right.", "---", "### Step 1: Simplify the Left Side Algebraically", "Subtract inside the parentheses carefully, distributing the negative sign:\n[\n(16a + 4b + c) - 4a - 2b - c\n]", "Group like terms:\n[\n(16a - 4a) + (4b - 2b) + (c - c) = 12a + 2b + 0\n]", "So the left side simplifies to:\n[\n12a + 2b\n]", "---", "### Step 2: Simplify the Right Side", "[\n15 - 3 = 12\n]", "---", "### Step 3: Equate Both Simplified Sides", "Now substitute simplified expressions:\n[\n12a + 2b = 12\n]", "This is our first derived equation.", "---", "### Step 4: Simplify Further (Optional but Insightful)", "Divide both sides by 6:\n[\n\frac{12a + 2b}{6} = \frac{12}{6} \Rightarrow 2a + \frac{1}{3}b = 2\n]", "Alternatively, divide by 2:\n[\n6a + b = 6\n]", "This is your second simplified equation:\n[\n6a + b = 6\n]", "---", "### Why This Step Matters: Subtracting Second from Third", "This type of operation illustrates a foundational algebraic principle — evaluating expressions by subtraction to isolate variables and produce manageable equations. By computing the difference ( (16a + 4b + c) - (4a + 2b + c) ), we:", "- Eliminate redundant terms (like ( c - c = 0 ))\n- Reduce complexity by combining like terms\n- Form a clear linear relationship in ( a ) and ( b )\n- Progress toward solving for variables or substituting in systems", "---", "### Summary of Key Result", "From the transaction:\n[\n(16a + 4b + c) - (4a + 2b + c) = 15 - 3\n]\nwe simplified both sides:\n- Left: ( 12a + 2b )\n- Right: ( 12 )\nEquating gives:\n[\n12a + 2b = 12 \Rightarrow 6a + b = 6\n]", "This methodically breaks complex expressions into simpler forms, enabling clearer problem-solving in algebra, systems of equations, and even advanced calculus applications.", "---", "SEO Keywords:\nSubtracting expressions algebraically, Simplify linear expressions, Solve equations step-by-step, Algebraic manipulation, Linear equation derivation, Algebra tutorial, Variable elimination, Solve for variables, Algebraic expressions example, Solve ( 6a + b = 6 )", "---", "By mastering such operations, students and learners strengthen their algebraic foundations, making advanced math tasks easier and more intuitive."]