Solve for t in the equation \(\frac{3}{4}\)t + \(\frac{1}{3}\)(21 – t) = 11

Explanation

\(\frac{3}{4}\) t + \(\frac{1}{3}\) (21 - t) = 11

  Multiply through by the LCM of 4 and 3 which is 12

  12 x(\(\frac{3}{4}\) t) + 12 x (\(\frac{1}{3}\) (21 - t)) = (11 x 12)

  9t + 4(21 - t) = 132

  9t + 84 - 4t = 132

  5t + 84 = 132

  5t = 132 - 84 = 48

  t = \(\frac{48}{5}\)

  t = 9 \(\frac{3}{5}\)

  Answer is D

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