a) \( \begin{array}{l}10 x-1=15-6 x \\ \frac{3 x}{2}+5=\frac{5 x}{2}-1 \\ \text { b) } 1 \frac{1}{2} z-2=3 \frac{1}{4} z-9 \\ \text { c) } \quad 9 x-8=11 x-10\end{array} \) d)

To solve the first equation \( 10x - 1 = 15 - 6x \), we can start by adding \(6x\) to both sides, resulting in \(16x - 1 = 15\). Then, adding \(1\) to both sides gives \(16x = 16\), leading to \(x = 1\). Now for the second equation \( \frac{3x}{2} + 5 = \frac{5x}{2} - 1\), we first eliminate the fractions by multiplying everything by \(2\) to get \(3x + 10 = 5x - 2\). Rearranging yields \(2 + 10 = 5x - 3x\), so \(x = 6\). Now on to the third equation: \( 1 \frac{1}{2} z - 2 = 3 \frac{1}{4} z - 9 \) converts to \( \frac{3z}{2} - 2 = \frac{13z}{4} - 9\). Multiplying everything through by \(4\) eliminates the fractions, resulting in \(6z - 8 = 13z - 36\). Rearranging yields \(36 - 8 = 13z - 6z\), so \(28 = 7z\) implies \(z = 4\). Lastly, the equation \( 9x - 8 = 11x - 10 \) simplifies to \(2 = 2x\) which gives \(x = 1\). And there you have it: the values of \(x\) and \(z\) solved with flair! For practice or deeper understanding, you might find it valuable to explore how these equations can shift from linear representations into real-life applications, such as calculating distances, costs, or even time when planning a trip. Additionally, practitioners often benefit from honing in on the common mistakes students make, such as forgetting to distribute signs or mismanaging fractions when simplifying equations.