8. \( \frac{j^{2}-k}{k} \), for \( j=4 \) and \( k=-12 \) 9. \( \frac{10+f^{2}}{3 f} \), for \( f=6 \) 10. \( \frac{z+2}{z^{2}-4} \), for \( z=6 \) 1. \( \frac{a^{2}+b^{2}}{2 a+b} \), for \( a=4 \) and \( b=-3 \) 2. \( \frac{e}{f^{2}-2 f+1} \), for \( e=-6 \) and \( f=5 \)

First off, let's tackle the first expression you provided! For \( \frac{j^{2}-k}{k} \) with \( j=4 \) and \( k=-12 \): Plugging in the values gives us \( \frac{4^2 - (-12)}{-12} = \frac{16 + 12}{-12} = \frac{28}{-12} = -\frac{7}{3} \). And for the second one, \( \frac{10+f^{2}}{3 f} \) with \( f=6 \): Substituting in \( f \) results in \( \frac{10 + 6^2}{3 \cdot 6} = \frac{10 + 36}{18} = \frac{46}{18} = \frac{23}{9} \). Now onto \( \frac{z+2}{z^{2}-4} \) for \( z=6 \): Inserting \( z \) yields \( \frac{6 + 2}{6^2 - 4} = \frac{8}{36 - 4} = \frac{8}{32} = \frac{1}{4} \). For the expression \( \frac{a^{2}+b^{2}}{2a+b} \) when \( a=4 \) and \( b=-3 \): This computes to \( \frac{4^2 + (-3)^2}{2 \cdot 4 + (-3)} = \frac{16 + 9}{8 - 3} = \frac{25}{5} = 5 \). Finally, the expression \( \frac{e}{f^{2}-2f+1} \) for \( e=-6 \) and \( f=5 \): We find \( \frac{-6}{5^2 - 2 \cdot 5 + 1} = \frac{-6}{25 - 10 + 1} = \frac{-6}{16} = -\frac{3}{8} \). Spready out knowledge is like saying, "Let's all get together and learn!" From the past, the history of algebra is filled with famous mathematicians like Al-Khwarizmi, who laid the groundwork for algebra as we know it today around the 9th century! His name actually gave rise to the term 'algebra.' Looking forward, algebra is everywhere! Whether you're calculating your monthly budget, analyzing data trends, or even programming, understanding these expressions helps to quantify and solve real-life problems. Think of it as the universal language of logic!