Solving equations 1 and 2: From 1, $a = 5 - 3b$. Substitute into 2: $2(5 - 3b) + 4b = 6 \Rightarrow 10 - 6b + 4b = 6 \Rightarrow -2b = -4 \Rightarrow b = 2$, then $a = 5 - 6 = -1$.

Solving equations 1 and 2: From 1, $a = 5 - 3b$. Substitute into 2: $2(5 - 3b) + 4b = 6 \Rightarrow 10 - 6b + 4b = 6 \Rightarrow -2b = -4 \Rightarrow b = 2$, then $a = 5 - 6 = -1$.

["Solving Equations 1 and 2: A Step-by-Step Guide to Finding $ a $ and $ b $", "Solving systems of equations is a fundamental skill in algebra, essential for students and professionals alike. Whether dealing with linear equations in two variables or more complex models, understanding substitution is key to finding precise solutions. In this article, we’ll walk through a clear, practical example: solving equations 1 and 2 using substitution to determine the values of $ a $ and $ b $. By mastering the steps shown, you’ll gain confidence in solving similar algebraic problems.", "---", "### The Problem: Solving Two Equations", "We are given the following system:", "- Equation 1:\n $$ a = 5 - 3b $$\n- Equation 2:\n $$ 2(5 - 3b) + 4b = 6 $$", "Our goal is to solve this system step-by-step to find $ a $ and $ b $.", "---", "### Step 1: Understand Equation 1 — Express $ a $ in terms of $ b $", "Equation 1 is already solved for $ a $:\n$$ a = 5 - 3b $$\nThis expression shows that the value of $ a $ depends directly on $ b $, making it ideal to substitute into Equation 2. The first step in solving paired equations is often isolating one variable, which we’ve done here.", "---", "### Step 2: Substitute $ a $ into Equation 2", "Equation 2 contains $ (5 - 3b) $, which we substitute from Equation 1:\n$$ 2(5 - 3b) + 4b = 6 $$", "Now, replace $ a $ in Equation 2:\nNote: Although Equation 1 gives $ a $, Equation 2 involves only $ a $ and $ b $, so substituting here focuses on solving for $ b $ alone.", "---", "### Step 3: Simplify and Solve for $ b $", "Expand the expression:\n$$ 2(5 - 3b) + 4b = 6 $$\n$$ 10 - 6b + 4b = 6 $$\nCombine like terms:\n$$ 10 - 2b = 6 $$\nSubtract 10 from both sides:\n$$ -2b = 6 - 10 $$\n$$ -2b = -4 $$\nDivide both sides by $-2$:\n$$ b = 2 $$", "---", "### Step 4: Back-Substitute to Find $ a $", "Now that we have $ b = 2 $, substitute this back into Equation 1:\n$$ a = 5 - 3b = 5 - 3(2) = 5 - 6 = -1 $$", "---", "### Final Answer", "The solution to the system is:\n$$\na = -1, \quad b = 2\n$$", "This means when $ b = 2 $, the corresponding value of $ a $ that satisfies both equations is $-1$.", "---", "### Why This Method Works", "Using substitution turns a system of two equations into a single equation with one variable—simpler to solve. Solving for one variable in terms of the other, then plugging into the second equation, avoids the complexity of eliminating variables algebraically. This method applies not only to linear equations but also to more advanced systems in algebra, calculus, and applied mathematics.", "---", "### Practical Tips for Solving Equations 1 and 2", "- Always verify substitutions: Check that each expression matches the original equation.\n- Combine like terms carefully to avoid errors in simplification.\n- Use inverse operations when isolating variables—subtract first, then divide.\n- Double-check your final values by plugging $ a $ and $ b $ back into both equations.", "---", "### Summary", "Solving equations 1 and 2 by substitution demonstrates a reliable algebraic technique:\nIsolate a variable, substitute into the other equation, simplify, then solve.\nThis process leads to $ b = 2 $ and $ a = -1 $, confirming the solution matches system constraints perfectly.", "Whether in homework, exams, or real-world modeling, mastering these steps builds a strong foundation in problem-solving with equations.", "---", "Keywords: solve equations, substitution method, linear equations, algebra tutorial, solve for $ a $ and $ b $, step-by-step, mathematical problem solving, equation substitution, algebraic techniques."]

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