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NECO 2004 Mathematics · Question 22

Find the value of t such that the expression 1t\frac{1}{t} + 43\frac{4}{3} - 56t\frac{5}{6t} + 1 is equal to zero

Options

  1. A.16\frac{1}{6}
  2. B.1−14\frac{1}{-14}
  3. C.- 32\frac{3}{2}
  4. D.76\frac{7}{6}
  5. E.25\frac{2}{5}

Answer: B. 1−14\frac{1}{-14}

Explanation written with AI and checked against the answer key

Analyze the question: We need t such that 1/t + 4/3 − 5/(6t) + 1 = 0. Concept: Group the terms with t in the denominator together, group the constants together, then solve. Solution: Terms with t: 1/t − 5/(6t) = 6/(6t) − 5/(6t) = 1/(6t) Constants: 4/3 + 1 = 7/3 So the equation becomes: 1/(6t) + 7/3 = 0 1/(6t) = −7/3 Cross-multiply: 3 = −42t t = −3/42 = −1/14 Check: If t = −1/14, then 1/t = −14 and 5/(6t) = 5/(−6/14) = −70/6 = −35/3. −14 + 4/3 + 35/3 + 1 = −13 + 39/3 = −13 + 13 = 0 ✓ Answer: t = −1/14, which is option B (written as 1/−14). Tip: In equations with fractions, first combine like terms (those with the same denominator variable). This reduces the work and the chance of errors. Always substitute back to verify.

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