[tex]\text{If is}\ -\dfrac{1}{6}b\ \text{then}\\\\-\dfrac{1}{6}b=3\qquad\text{multiply both sides by (-6)}\\\\(-6\!\!\!\!\!\diagup^1)\cdot\left(-\dfrac{1}{\not6_1}b\right)=3(-6)\\\\(-1)(-b)=-18\\\\\boxed{b=-18}\to\boxed{a.}\\\\\text{If is}\ -16b,\ \text{then}\\\\-16b=3\qquad\text{divide both sides by (-16)}\\\\b=\dfrac{3}{-16}\\\\\boxed{b=-\dfrac{1}{16}}[/tex]
