We solve this using elimination. Multiply the first equation by 2 and the second by 5:
["Solving Systems of Equations Using Elimination: A Step-by-Step Guide", "When tackling systems of linear equations, the elimination method offers a powerful and efficient strategy—especially when combined with smart manipulation like multiplying equations to align coefficients. One common and effective approach involves multiplying the first equation by 2 and the second by 5, aligning key variables to simplify elimination.", "### Why Use the Elimination Method?", "The elimination method lets you eliminate one variable by combining equations, making it easier to solve for the remaining unknowns. Unlike substitution, which often requires isolating terms first, elimination streamlines the process by leveraging coefficients to cancel variables cleanly.", "This technique shines in word problems, real-world modeling, and advanced math competitions where clarity and precision matter.", "### How It Works: Multiplying First by 2, Second by 5", "Let’s explore a typical scenario:\nSolve the system:\n[\n\begin{cases}\na + b = 7 \\n2a + 5b = 29\n\end{cases}\n]", "Step 1: Align Coefficients Through Multiplication\nTo eliminate either ( a ) or ( b ), multiply equations to make coefficients opposites. Here, multiply the first equation by 2:\n[\n2(a + b) = 2 \cdot 7 \Rightarrow 2a + 2b = 14\n]\nAnd multiply the second equation by 5:\n[\n5(2a + 5b) = 5 \cdot 29 \Rightarrow 10a + 25b = 145\n]", "Now your system looks like:\n[\n\begin{cases}\n2a + 2b = 14 \\n10a + 25b = 145\n\end{cases}\n]", "Step 2: Eliminate One Variable\nSubtract ( 5 \ imes ) the first new equation from the second to eliminate ( a ):\n[\n(10a + 25b) - 5(2a + 2b) = 145 - 5 \cdot 14\n\Rightarrow 10a + 25b - 10a - 10b = 145 - 70\n\Rightarrow 15b = 75\n\Rightarrow b = 5\n]", "Step 3: Back-Substitute to Find the Other Variable\nPlug ( b = 5 ) into the first original equation:\n[\na + 5 = 7 \Rightarrow a = 2\n]", "### Final Answer", "The solution is ( a = 2 ), ( b = 5 ). This method works efficiently when equations are designed for easy coefficient alignment.", "### When to Use This Tactic\n- Systems where variables have integer coefficients that allow simple multiplication\n- Word problems requiring quick setup and elegant elimination\n- Enhancing problem-solving speed in academic settings or real-world analytics", "### Key Takeaway", "Multiplying equations before elimination transforms complex systems into manageable forms. By strategically scaling terms—like multiplying the first equation by 2 and the second by 5—we create matching coefficients that cancel one variable instantly. This approach not only saves time but also reduces errors, making it a cornerstone skill in algebra and beyond.", "---", "Try it yourself with your own equations—multiplying first by 2 and then by 5—and watch how elimination becomes second nature!"]
