Solve for \( d \).
\[
\frac{1}{6}d + \frac{2}{3} = \frac{1}{4}(d - 2)
\]
Answer (2)
\frac{1}{6}d+\frac{2}{3}=\frac{1}{4}(d-2) \\ \\ \frac{d}{6}^{(2}}+\frac{2}{3}^{(4}=\frac{d-2}{4}^{(3} \\ \\ 2d+8=3(d-2) \\ \\ 2d+8=3d-6 \\ \\ 2d-3d=-6-8 \\ \\ -d=-14 \\ \\ \boxed{d=14}
To solve the equation 6 1 d + 3 2 = 4 1 ( d − 2 ) , we first eliminate fractions by multiplying by 12, leading to the simplified equation 2 d + 8 = 3 d − 6 . Rearranging gives d = 14 .
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