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QUESTION 4. Write the following as simply as possible without using negative indices and as a single fraction.

Mathematics

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Answers

2 parts
a

8x10y2\frac{8}{x^{10}y^2}

b

xy2+22y3\frac{xy^2 + 2}{2y^3}

Step-by-step solution

11 steps

Let's re-examine Question 4, specifically part b), as there was a misread operator in the previous solution.

Part 4

Write the following as simply as possible without using negative indices and as a single fraction.

Part a

8x10y2\frac{8}{x^{10}y^2}

(2x−3)3y−2x\frac{(2x^{-3})^3 y^{-2}}{x}

  1. Apply the power rule (ab)n=anbn(ab)^n = a^n b^n and (am)n=amn(a^m)^n = a^{mn} to the numerator.

    (2x−3)3=23(x−3)3=8x−9(2x^{-3})^3 = 2^3 (x^{-3})^3 = 8x^{-9}

  2. Substitute this back into the expression.

    8x−9y−2x\frac{8x^{-9} y^{-2}}{x}

  3. Simplify using the division rule am÷an=am−na^m \div a^n = a^{m-n}.

    8x−9y−2x1=8x−9−1y−2=8x−10y−2\frac{8x^{-9} y^{-2}}{x^1} = 8x^{-9-1} y^{-2} = 8x^{-10} y^{-2}

  4. Rewrite the expression without negative indices and as a single fraction using a−n=1ana^{-n} = \frac{1}{a^n}.

    8x−10y−2=8x10y28x^{-10} y^{-2} = \frac{8}{x^{10}y^2}

    The final answer is 8x10y2\boxed{\frac{8}{x^{10}y^2}}.

Part b

xy2+22y3\frac{xy^2 + 2}{2y^3}

x(2−1y+y−1x−1)÷(xy)2x2x(2^{-1}y + y^{-1}x^{-1}) \div \frac{(xy)^2}{x^2}

  1. Simplify the terms with negative exponents inside the parenthesis.

    2−1=122^{-1} = \frac{1}{2} y−1=1yy^{-1} = \frac{1}{y} x−1=1xx^{-1} = \frac{1}{x}

    The expression inside the parenthesis becomes 12y+1y1x=y2+1xy\frac{1}{2}y + \frac{1}{y}\frac{1}{x} = \frac{y}{2} + \frac{1}{xy}.

  2. Distribute xx into the parenthesis.

    x(y2+1xy)=x⋅y2+x⋅1xy=xy2+1yx\left(\frac{y}{2} + \frac{1}{xy}\right) = x \cdot \frac{y}{2} + x \cdot \frac{1}{xy} = \frac{xy}{2} + \frac{1}{y}

  3. Simplify the divisor term.

    (xy)2x2=x2y2x2=y2\frac{(xy)^2}{x^2} = \frac{x^2 y^2}{x^2} = y^2

  4. Perform the division.

    The expression is now (xy2+1y)÷y2\left(\frac{xy}{2} + \frac{1}{y}\right) \div y^2.
    To divide by y2y^2, multiply by its reciprocal, 1y2\frac{1}{y^2}.

    (xy2+1y)×1y2\left(\frac{xy}{2} + \frac{1}{y}\right) \times \frac{1}{y^2}

  5. Distribute 1y2\frac{1}{y^2} into the parenthesis.

    xy2×1y2+1y×1y2=x2y+1y3\frac{xy}{2} \times \frac{1}{y^2} + \frac{1}{y} \times \frac{1}{y^2} = \frac{x}{2y} + \frac{1}{y^3}

  6. Find a common denominator and combine the fractions.

    The least common multiple of 2y2y and y3y^3 is 2y32y^3.

    x⋅y22y⋅y2+1⋅2y3⋅2=xy22y3+22y3\frac{x \cdot y^2}{2y \cdot y^2} + \frac{1 \cdot 2}{y^3 \cdot 2} = \frac{xy^2}{2y^3} + \frac{2}{2y^3}

  7. Add the fractions.

    xy2+22y3\frac{xy^2 + 2}{2y^3}

    The final answer is xy2+22y3\boxed{\frac{xy^2 + 2}{2y^3}}.

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