\begin{tabular}{|l|l|}\hline & O The system has no solution. \\ \( \begin{array}{l}\text { System B } \\ 4 x-y=8 \\ -4 x+y=-8\end{array} \) & \( \begin{array}{l}\text { The system has a unique solution: } \\ \text { The system has infinitely many solutions. } \\ \text { They must satisfy the following equation: } \\ y=\square\end{array} \)\end{tabular}

b. \( -0.5 p+\frac{1}{2} p-2.75+\frac{2}{3} \)

Consider the line \( y=\frac{5}{2} x-3 \). Find the equation of the line that is perpendicular to this line and passes through the point \( (2,-5) \). Find the equation of the line that is parallel to this line and passes through the point (2, -5). Note that the ALEKS graphing calculator may be helpful in checking your answer. Equation of perpendicular line:

Use the properties of logarithms to rewrite the expression. Simplify the res \( \log _{3} \frac{\sqrt{x} \sqrt[3]{y}}{w} \sqrt{z} \) Choose the correct answer. A. \( -\frac{1}{2} \log _{3} x-\frac{1}{3} \log _{3} y+5 \log _{3} w+\frac{1}{2} \log _{3} z \) B. \( \frac{1}{2} \log _{3} x+\frac{1}{3} \log _{3} y-5 \log _{3} w-\frac{1}{2} \log _{3} z \) C. \( { }^{2} \log _{3} x+3 \log _{3} y-5 \log _{3} w-2 \log _{3} z \) D. This cannot be simplified.

Solve the equation by factoring. Select the correct solution. \( 6 x^{2}-5 x-4=0 \) \( \begin{array}{ll}\text { O } & \text { A) } x=-\frac{4}{3},-\frac{1}{2} \\ \text { O } & \text { B) } x=-\frac{1}{2}, \frac{4}{3} \\ \text { O } & \text { C) } x=-\frac{4}{3}, \frac{1}{2} \\ \text { - } & \text { D) } x=\frac{1}{2}, \frac{4}{3}\end{array} \)

1 Graph \( y-2=\frac{2}{3}(x-3) \)