[tex]\dfrac{5}{3}r+\dfrac{4}{3}=-\dfrac{8}{9}\ \ \ \ |multiply\ both\ sides\ by\ 9\\\\3\cdot5r+3\cdot4=-8\\\\15r+12=-8\ \ \ \ |subtract\ 12\ from\ both\ sides\\\\15r=-20\ \ \ \ |divide\ both\ sides\ by\ 15\\\\r=-\dfrac{20}{15}\\\\r=-\dfrac{4}{3}[/tex]
