Question

Simplify the following


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Answer 1

Answer:

$\dfrac{1-\dfrac{4}{z}-\dfrac{21}{z^2}}{1-\dfrac{2}{z}-\dfrac{15}{z^2}}=\dfrac{z-7}{z-5}$

Step-by-step explanation:

$\dfrac{1-\dfrac{4}{z}-\dfrac{21}{z^2}}{1-\dfrac{2}{z}-\dfrac{15}{z^2}}$

$\gray{\mathrm{Join}\:1-\dfrac{2}{z}-\dfrac{15}{z^2}:\quad \dfrac{z^2-2z-15}{z^2}}$

$=\dfrac{1-\dfrac{4}{z}-\dfrac{21}{z^2}}{\dfrac{z^2-2z-15}{z^2}}$

$\gray{\mathrm{Join}\:1-\dfrac{4}{z}-\dfrac{21}{z^2}:\quad \dfrac{z^2-4z-21}{z^2}}$

$=\dfrac{\dfrac{z^2-4z-21}{z^2}}{\dfrac{z^2-2z-15}{z^2}}$

$\gray{\mathrm{Divide\:fractions}:\quad \dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a\cdot \:d}{b\cdot \:c}}$

$=\dfrac{\left(z^2-4z-21\right)z^2}{z^2\left(z^2-2z-15\right)}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:z^2}$

$=\dfrac{z^2-4z-21}{z^2-2z-15}$

$\gray{\mathrm{Factor}\:z^2-4z-21:\quad \left(z+3\right)\left(z-7\right)}$

$=\dfrac{\left(z+3\right)\left(z-7\right)}{z^2-2z-15}$

$\gray{\mathrm{Factor}\:z^2-2z-15:\quad \left(z+3\right)\left(z-5\right)}$

$=\dfrac{\left(z+3\right)\left(z-7\right)}{\left(z+3\right)\left(z-5\right)}$

$\gray{\mathrm{Cancel\:the\:common\:factor:}\:z+3}$

$=\dfrac{z-7}{z-5}$