\[ 2(3w + 5) = 58 \] - AIKO, infinite ways to autonomy.
Solving the Equation: 2(3w + 5) = 58 – A Step-by-Step Guide
Solving the Equation: 2(3w + 5) = 58 – A Step-by-Step Guide
Mastering algebra is essential for students, educators, and anyone looking to strengthen problem-solving skills. One of the most common equation types involves distributing, combining terms, and isolating variables — and the equation 2(3w + 5) = 58 is a perfect example. In this SEO-friendly article, we’ll break down how to solve this equation clearly, accurately, and using keywords that boost visibility for learners seeking algebra help.
Understanding the Context
What is the Equation We’re Solving?
We begin with:
2(3w + 5) = 58
This equation tells us that twice the quantity (3w + 5) equals 58. To find the value of the variable w, we follow a standard sequence of operations: distributing, simplifying, isolating the variable, and solving.
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Step-by-Step Solution
Step 1: Distribute the 2
Multiply 2 with both terms inside the parentheses:
2 × (3w + 5) = 2 × 3w + 2 × 5
→ 6w + 10 = 58
This step eliminates parentheses and prepares the equation for simplification.
Step 2: Subtract 10 from Both Sides
To isolate the term with the variable, subtract 10 from both sides:
6w + 10 – 10 = 58 – 10
→ 6w = 48
Step 3: Divide Both Sides by 6
Now divide both sides by 6 to solve for w:
6w ÷ 6 = 48 ÷ 6
→ w = 8
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Verification – Check the Solution
Plug w = 8 back into the original equation to confirm:
2(3(8) + 5) = 2(24 + 5) = 2 × 29 = 58 — matches perfectly!
Why This Equation Matters – Educational Relevance
Equations like 2(3w + 5) = 58 are foundational in algebra. They teach:
- Distributive property: Multiplying across parentheses
- Operations and balance: What you do to one side, do to the other
- Solution strategies: Isolating variables through inverse operations
Parents and students often search for clear, step-by-step algebra solutions to reinforce classroom learning — and understanding such equations helps build confidence and analytical thinking.
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