Problems 9-10: Solve each equation. Show your thinking. \( \begin{array}{ll}9 .-4(r+2)=4(2-2 r) & \text { 10. } 1.3+6 d=2.7-8 d\end{array} \)

Let's tackle the first equation: For the equation \(-4(r+2)=4(2-2r)\), we begin by distributing on both sides. The left side becomes \(-4r - 8\) and the right side becomes \(8 - 8r\). Now we have: \[ -4r - 8 = 8 - 8r \] Next, we can move all the \(r\) terms to one side and the constant terms to the other. Adding \(8r\) to both sides results in: \[ 4r - 8 = 8 \] Now, add \(8\) to both sides: \[ 4r = 16 \] Finally, divide by \(4\): \[ r = 4 \] For the second equation, \(1.3 + 6d = 2.7 - 8d\), we start by isolating the \(d\) terms. Let's add \(8d\) to both sides: \[ 1.3 + 14d = 2.7 \] Now, we subtract \(1.3\) from both sides: \[ 14d = 2.7 - 1.3 \] Calculating the right side gives us \(1.4\): \[ 14d = 1.4 \] Finally, divide both sides by \(14\): \[ d = \frac{1.4}{14} = 0.1 \] To summarize, the solutions are \(r = 4\) for the first equation and \(d = 0.1\) for the second!