Answer:
[tex]\large\boxed{\text{If is "or", then}\ x>-2\dfrac{1}{15}}\\\boxed{\text{If is "and", then}\ x>1\dfrac{9}{14}}[/tex]
Step-by-step explanation:
[tex]7x+\dfrac{3}{2}>13\qquad\text{multiply both sides by 2}\\\\14x+3>26\qquad\text{subtract 3 from both sides}\\\\14x>23\qquad\text{divide both sides by 14}\\\\x>\dfrac{23}{14}=1\dfrac{9}{14}\\=====================\\\\\dfrac{5}{2}x-\dfrac{1}{3}>-\dfrac{11}{2}\qquad\text{multiply both sides by 2}\\\\5x-\dfrac{2}{3}>-11\qquad\text{multiply both sides by 3}\\\\15x-2>-33\qquad\text{add 2 to both sides}\\\\15x>-31\qquad\text{divide both sides by 15}\\\\x>-\dfrac{31}{15}=-2\dfrac{1}{15}\\=====================[/tex]
