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Simplify the following expressions: 5d) 3/(2x3) - sqrt((8x2y3)/(2y)); 5f) 3/(2x3) - sqrt((32y8)/(9y2)) + sqrt((98y6)/(2))

Mathematics

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here are the solutions for questions 5d and 5f, simplified for a Grade 9 learner.

5d)
The expression is:

54x6y32x33−8x2y32y\sqrt[3]{\frac{54x^6y^3}{2x^3}} - \sqrt{\frac{8x^2y^3}{2y}}
  1. Simplify the fractions inside each radical.

    For the first term:

    54x6y32x3=542⋅x6x3⋅y3=27x6−3y3=27x3y3\frac{54x^6y^3}{2x^3} = \frac{54}{2} \cdot \frac{x^6}{x^3} \cdot y^3 = 27x^{6-3}y^3 = 27x^3y^3

    For the second term:

    8x2y32y=82⋅x2⋅y3y=4x2y3−1=4x2y2\frac{8x^2y^3}{2y} = \frac{8}{2} \cdot x^2 \cdot \frac{y^3}{y} = 4x^2y^{3-1} = 4x^2y^2

    Now the expression looks like:

    27x3y33−4x2y2\sqrt[3]{27x^3y^3} - \sqrt{4x^2y^2}
  2. Take the cube root and the square root.

    For the first term, 27x3y33\sqrt[3]{27x^3y^3}:

    273⋅x33⋅y33=3⋅x⋅y=3xy\sqrt[3]{27} \cdot \sqrt[3]{x^3} \cdot \sqrt[3]{y^3} = 3 \cdot x \cdot y = 3xy

    For the second term, 4x2y2\sqrt{4x^2y^2}:

    4⋅x2⋅y2=2⋅x⋅y=2xy\sqrt{4} \cdot \sqrt{x^2} \cdot \sqrt{y^2} = 2 \cdot x \cdot y = 2xy
  3. Combine the simplified terms.

    3xy−2xy=xy3xy - 2xy = xy

    The final answer is xy\boxed{xy}.

    5f)
    The expression is:

    16x3y92x33−32y89y2+98y62\sqrt[3]{\frac{16x^3y^9}{2x^3}} - \sqrt{\frac{32y^8}{9y^2}} + \sqrt{\frac{98y^6}{2}}
  4. Simplify the fractions inside each radical.

    For the first term:

    16x3y92x3=162⋅x3x3⋅y9=8y9\frac{16x^3y^9}{2x^3} = \frac{16}{2} \cdot \frac{x^3}{x^3} \cdot y^9 = 8y^9

    For the second term:

    32y89y2=329⋅y8y2=32y69\frac{32y^8}{9y^2} = \frac{32}{9} \cdot \frac{y^8}{y^2} = \frac{32y^6}{9}

    For the third term:

    98y62=982⋅y6=49y6\frac{98y^6}{2} = \frac{98}{2} \cdot y^6 = 49y^6

    Now the expression looks like:

    8y93−32y69+49y6\sqrt[3]{8y^9} - \sqrt{\frac{32y^6}{9}} + \sqrt{49y^6}
  5. Take the cube root and square roots.

    For the first term, 8y93\sqrt[3]{8y^9}:

    83⋅y93=2⋅y9/3=2y3\sqrt[3]{8} \cdot \sqrt[3]{y^9} = 2 \cdot y^{9/3} = 2y^3

    For the second term, 32y69\sqrt{\frac{32y^6}{9}}:

    32y69=16⋅2⋅y63=16⋅2⋅y63=42y33\frac{\sqrt{32y^6}}{\sqrt{9}} = \frac{\sqrt{16 \cdot 2 \cdot y^6}}{3} = \frac{\sqrt{16} \cdot \sqrt{2} \cdot \sqrt{y^6}}{3} = \frac{4\sqrt{2}y^3}{3}

    For the third term, 49y6\sqrt{49y^6}:

    49⋅y6=7⋅y6/2=7y3\sqrt{49} \cdot \sqrt{y^6} = 7 \cdot y^{6/2} = 7y^3
  6. Combine the simplified terms.

    2y3−42y33+7y32y^3 - \frac{4\sqrt{2}y^3}{3} + 7y^3

    Combine the terms with y3y^3:

    (2y3+7y3)−42y33=9y3−42y33(2y^3 + 7y^3) - \frac{4\sqrt{2}y^3}{3} = 9y^3 - \frac{4\sqrt{2}y^3}{3}

    To combine these, find a common denominator for the coefficients:

    9⋅33y3−423y3=273y3−423y3\frac{9 \cdot 3}{3}y^3 - \frac{4\sqrt{2}}{3}y^3 = \frac{27}{3}y^3 - \frac{4\sqrt{2}}{3}y^3 =27−423y3= \frac{27 - 4\sqrt{2}}{3}y^3

    The final answer is y3(27−42)3\boxed{\frac{y^3(27 - 4\sqrt{2})}{3}}.

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