Solve the following equations:
15) $8n + 7 = 31$
17) $\frac{n + 5}{-16} = -1$
The problem asks us to solve two separate linear equations for the variable \(n\).
1) \(8n + 7 = 31\)
2) \(\frac{n + 5}{-16} = -1\)
For the first equation, we need to isolate the variable \(n\). This involves undoing the operations performed on \(n\) in reverse order of operations. We start by subtracting the constant term from both sides of the equation.
For the second equation, we have a fraction. We can eliminate the fraction by multiplying both sides by the denominator, and then isolate the variable \(n\) by subtracting the remaining constant.
To solve the first equation \(8n + 7 = 31\):
\(8n = 31 - 7\)
\(8n = 24\)
\(n = \frac{24}{8}\)
\(n = 3\)
The solution to the first equation is \(n = 3\).
To solve the second equation \(\frac{n + 5}{-16} = -1\):
\(n + 5 = -1 \times (-16)\)
\(n + 5 = 16\)
\(n = 16 - 5\)
\(n = 11\)
The solution to the second equation is \(n = 11\).
Linear equations like these are fundamental in everyday life for solving simple allocation or budgeting problems. For example, if you are saving money to buy a video game costing \$31 and you already have \$7, you might solve \(8n + 7 = 31\) where \(8n\) represents the total money you save over \(n\) weeks, helping you determine how many weeks it takes to reach your goal. Similarly, solving \(\frac{n + 5}{-16} = -1\) could represent a scenario where a temperature or altitude change over a certain number of steps needs to be determined to reach a target value. Understanding these basic algebraic manipulations allows you to model and solve such problems efficiently.