14. Given:
[tex]\frac{3}{2}r-rs+4,r=\frac{6}{7},s=\frac{2}{3}[/tex]
Substitute the value of r and s in the given problem.
We get,
[tex]\begin{gathered} \frac{3}{2}(\frac{6}{7})-(\frac{6}{7})(\frac{2}{3})+4=3(\frac{3}{7})-(\frac{2}{7})(2)+4 \\ =\frac{9}{7}-\frac{4}{7}+4 \\ =\frac{5}{7}+4 \\ =\frac{33}{7} \end{gathered}[/tex]
Hence, the answer is
[tex]\frac{33}{7}[/tex]
