\( \left.\begin{array}{l}\text { Now, solve for the value of } b \text {, 4) } \\ \qquad \begin{array}{r}s=4 b \\ s+b=55\end{array} \\ s+b=55 \\ 4 b+b=55\end{array}\right\} \begin{array}{r}4 b+b \quad \begin{array}{r}55\end{array}\end{array} \)

To solve for \( b \), you can first combine the two equations. From the first equation, \( s = 4b \). Substituting this into the second equation results in: \[ 4b + b = 55 \] which simplifies to: \[ 5b = 55. \] Dividing both sides by 5 gives: \[ b = 11. \] Now you know the value of \( b \) is 11! Have you ever thought about how such equations can model real-life situations? For instance, in business, we often deal with linear relationships like this one when determining costs, profits, or resources. If \( s \) represented the total salary in a job scenario and \( b \) a bonus structure, finding out how bonuses relate to total compensation is crucial for making sound financial decisions. It's all about understanding the variables at play! Also, it's easy to make small errors in algebra, especially during substitutions! Always double-check that when you substitute, you are putting in the correct expressions and keeping track of any signs. Misplacing a negative or forgetting to distribute can lead to a wrong answer. Keeping your equations organized will help avoid these common pitfalls!