(a) Evaluate : \(2 \div (\frac{64}{125})^{-\frac{2}{3}}\) (b) The lines \(y = 3x + 5\) and \(y = - 4x - 1\) intersect at a point k. Find the coordinates of ...

Answer Details

(a) A negative index inverts the base:

\[\left(\tfrac{64}{125}\right)^{-\frac{2}{3}} = \left(\tfrac{125}{64}\right)^{\frac{2}{3}} = \left(\sqrt[3]{\tfrac{125}{64}}\right)^{2} = \left(\tfrac{5}{4}\right)^{2} = \tfrac{25}{16}.\]

\[2 \div \tfrac{25}{16} = 2 \times \tfrac{16}{25} = \tfrac{32}{25} = 1\tfrac{7}{25}.\]

(b) At the point of intersection k the two y-values are equal:

\[3x + 5 = -4x - 1 \Rightarrow 7x = -6 \Rightarrow x = -\tfrac{6}{7}.\]

\[y = 3\left(-\tfrac{6}{7}\right) + 5 = -\tfrac{18}{7} + \tfrac{35}{7} = \tfrac{17}{7}.\]

Coordinates of k: \(\left(-\tfrac{6}{7},\ \tfrac{17}{7}\right)\).