Verified Answer :

  • Distribute the constant on the left side: \(5n + 15 = 3n + 1\).
  • Subtract \(3n\) from both sides to group variable terms: \(2n + 15 = 1\).
  • Subtract 15 from both sides to isolate the variable term: \(2n = -14\).
  • Divide by 2 to solve for \(n\): \(n = \boxed{-7}\).

Explanation

  1. Analysis of the Equation

We are given the equation \(5(n+3) = 3n + 1\). This is a linear equation in one variable, \(n\). Our objective is to find the value of \(n\) that makes this equation true. We will do this by performing algebraic operations on both sides to isolate \(n\).

  1. Expanding the Equation

First, distribute the constant 5 on the left side of the equation:
\[5 \cdot n + 5 \cdot 3 = 3n + 1\]
\[5n + 15 = 3n + 1\]

  1. Grouping the Variable Terms

Next, we group all terms involving \(n\) on one side of the equation by subtracting \(3n\) from both sides:
\[5n - 3n + 15 = 1\]
\[2n + 15 = 1\]

  1. Isolating the Variable Term

To isolate the term with \(n\), subtract 15 from both sides of the equation:
\[2n = 1 - 15\]
\[2n = -14\]

  1. Solving for n

Finally, solve for \(n\) by dividing both sides of the equation by 2:
\[n = \frac{-14}{2}\]
\[n = -7\]
We can check the solution by substituting \(n = -7\) back into the original equation: \(5(-7 + 3) = 5(-4) = -20\) and \(3(-7) + 1 = -21 + 1 = -20\). Since both sides are equal, the solution is correct. The final value of \(n\) is \(\boxed{-7}\).

Examples

Solving linear equations like this is essential for balancing budgets. Imagine you have a subscription fee (the \(3n + 1\) part, where \(n\) is the number of months) and you are comparing it with an upfront lump sum plus a fixed monthly fee (the \(5(n+3)\) part). Understanding how to isolate variables helps you identify at what point these two costs become equivalent, allowing you to make smarter financial decisions.

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