eq 3 $. Set $ x + 3 = 5 $, so $ x = 2 $. Verify $ x = 2 $ does not violate the restriction. Final answer: $ oxed2 $.

Solving Equation 3: $ x + 3 = 5 $

Solving simple linear equations is a fundamental skill in algebra, and Equation 3 offers a perfect example of direct application and verification. Let’s walk through how to solve $ x + 3 = 5 $, confirm the solution, and discuss any restrictions that may apply.

Step-by-Step Solution

Understanding the Context

We begin with the equation:
$$
x + 3 = 5
$$

To isolate $ x $, subtract 3 from both sides:
$$
x + 3 - 3 = 5 - 3
$$

This simplifies to:
$$
x = 2
$$

Verification of the Solution

eq 3 $. Set $ x + 3 = 5 $, so $ x = 2 $. Verify $ x = 2 $ does not violate the restriction. Final answer: $ oxed{2} $.

Key Insights

It’s essential to verify that the found value satisfies the original equation without violating any constraints. Substitute $ x = 2 $ back into the left-hand side:
$$
x + 3 = 2 + 3 = 5
$$

The result matches the right-hand side, confirming the solution is correct.

Checking for Restrictions

Now, verify if $ x = 2 $ violates any restrictions. In this basic equation, no divisors, radicals, logarithms, or other operations introduce domain constraints. Since multiplication or division by zero is not involved, $ x = 2 $ is fully valid. There are no excluded values or algebraic restrictions that limit the solution.

Final Answer

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Final Thoughts

The solution to $ x + 3 = 5 $ is verified and unrestricted. Therefore:
$$
oxed{2}
$$