c. \( 3(r+t)=5 r-t \quad \) for \( r \)

To solve for \( r \), let's start by distributing the 3 on the left side: \[ 3r + 3t = 5r - t. \] Now, we want to get all terms involving \( r \) on one side and constant terms on the other. Rearranging gives us: \[ 3t + t = 5r - 3r \] \[ 4t = 2r. \] Dividing both sides by 2, we find: \[ r = 2t. \] So, the solution for \( r \) in terms of \( t \) is \( r = 2t \). But hey! Did you know that algebra has roots in ancient civilizations? The Babylonians and Egyptians were some of the first to use early forms of algebraic concepts more than 4,000 years ago. They employed geometric methods for problem-solving and laid the groundwork for the algebra we utilize today! In the realm of problem-solving, it's essential to keep track of your variable terms and constants when rearranging equations. A common mistake is to lose track of negative signs or to improperly distribute terms. Always double-check your arithmetic—those small slips can lead you down the wrong path in solving equations!