YES. She is correct.
[tex]\dfrac{x}{16}=4\qquad\text{multiply both sides by 4}\\\\\not\!4^1\cdot\dfrac{x}{16\!\!\!\!\!\diagup_4}=4\cdot4\\\\\dfrac{x}{4}=16[/tex]
Equations are equivalent
Their solutions:
[tex]\dfrac{x}{16}=4\qquad\text{multiply both sides by 16}\\\\x=4\cdot16\\\\\boxed{x=64}\\----------\\\\\dfrac{x}{4}=16\qquad\text{multiply both sides by 4}\\\\x=16\cdot4\\\\\boxed{x=64}[/tex]
