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Evaluate ((3)/(7) - (5)/(8)) x (2)/(3)3 + (1)/(7) of (2)/(3)

Mathematics

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1. Evaluate

(37−58)×233+17 of 23\frac{\left(\frac{3}{7} - \frac{5}{8}\right) \times \frac{2}{3}}{3 + \frac{1}{7} \text{ of } \frac{2}{3}}
  1. Evaluate the expression in the numerator.

    First, calculate the difference inside the parenthesis:

    37−58=3×8−5×77×8=24−3556=−1156\frac{3}{7} - \frac{5}{8} = \frac{3 \times 8 - 5 \times 7}{7 \times 8} = \frac{24 - 35}{56} = \frac{-11}{56}

    Next, multiply the result by 23\frac{2}{3}:

    −1156×23=−11×256×3=−22168=−1184\frac{-11}{56} \times \frac{2}{3} = \frac{-11 \times 2}{56 \times 3} = \frac{-22}{168} = \frac{-11}{84}

    So, the numerator is −1184\frac{-11}{84}.

  2. Evaluate the expression in the denominator.

    First, calculate 17 of 23\frac{1}{7} \text{ of } \frac{2}{3}:

    17×23=1×27×3=221\frac{1}{7} \times \frac{2}{3} = \frac{1 \times 2}{7 \times 3} = \frac{2}{21}

    Next, add 3 to the result:

    3+221=3×2121+221=6321+221=65213 + \frac{2}{21} = \frac{3 \times 21}{21} + \frac{2}{21} = \frac{63}{21} + \frac{2}{21} = \frac{65}{21}

    So, the denominator is 6521\frac{65}{21}.

  3. Divide the numerator by the denominator.

    −11846521=−1184×2165\frac{\frac{-11}{84}}{\frac{65}{21}} = \frac{-11}{84} \times \frac{21}{65}

    Simplify by dividing 84 by 21 (84÷21=484 \div 21 = 4):

    −114×65=−11260\frac{-11}{4 \times 65} = \frac{-11}{260}

    The evaluated expression is −11260\boxed{\frac{-11}{260}}.

    2. A two-digit number is such that when the digits are reversed, the value of the number increases by 36. If the sum of the unit digit and twice the tens digit is 16, find the number.

  4. Define the two-digit number using variables.

    Let the tens digit be tt and the unit digit be uu.
    The original number can be represented as 10t+u10t + u.
    The number with digits reversed is 10u+t10u + t.

  5. Formulate equations based on the given conditions.

    Condition 1: When the digits are reversed, the value of the number increases by 36.

    (10u+t)−(10t+u)=36(10u + t) - (10t + u) = 36 9u−9t=369u - 9t = 36

    Divide by 9:

    u−t=4(Equation 1)u - t = 4 \quad \text{(Equation 1)}

    Condition 2: The sum of the unit digit and twice the tens digit is 16.

    u+2t=16(Equation 2)u + 2t = 16 \quad \text{(Equation 2)}
  6. Solve the system of linear equations.

    From Equation 1, express uu in terms of tt:

    u=t+4u = t + 4

    Substitute this expression for uu into Equation 2:

    (t+4)+2t=16(t + 4) + 2t = 16 3t+4=163t + 4 = 16 3t=16−43t = 16 - 4 3t=123t = 12 t=123t = \frac{12}{3} t=4t = 4

    Now substitute the value of tt back into u=t+4u = t + 4:

    u=4+4u = 4 + 4 u=8u = 8
  7. Form the original number.

    The tens digit is t=4t=4 and the unit digit is u=8u=8.
    The number is $1

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