How do you multiply $(\frac{1}{m^{2} - m}) + (\frac{1}{m}) = \frac{5}{m^{2} - m}$ $(1/(m^2-m)) + (1/m)=5/(m^2-m)$ ?

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How do you multiply $(\frac{1}{m^{2} - m}) + (\frac{1}{m}) = \frac{5}{m^{2} - m}$ $(1/(m^2-m)) + (1/m)=5/(m^2-m)$ ?

1 Answer

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Answer 1 · 2018-04-11T21:49:03

$m = 5$ $m = 5$

Explanation:

the common denominator between $m^{2} - m$ $m^2 - m$ and $m$ $m$ would be $m^{2} - m$ $m^2 - m$ , since $m^{2} - m$ $m^2 - m$ is a multiple of $m$ $m$ .

$m^{2} - m = m (m - 1)$ $m^2 - m = m(m - 1)$

$\frac{1}{m} = \frac{1 (m - 1)}{m (m - 1)}$ $1/m = (1(m-1))/(m(m-1))$

$= \frac{m - 1}{m^{2} - m}$ $= (m - 1)/(m^2-m)$

$\frac{1}{m^{2} - m} + \frac{m - 1}{m^{2} - m} = \frac{1 + m - 1}{m^{2} - m}$ $1/(m^2-m) + (m-1)/(m^2-m) = (1 + m - 1)/(m^2-m)$

$= \frac{m}{m^{2} - m}$ $= m / (m^2 - m)$

$\frac{m}{m^{2} - m} = \frac{5}{m^{2} - m}$ $m / (m^2-m) = 5 / (m^2 - m)$

hence, $m = 5$ $m = 5$

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How do you multiply $(\frac{1}{m^{2} - m}) + (\frac{1}{m}) = \frac{5}{m^{2} - m}$ $(1/(m^2-m)) + (1/m)=5/(m^2-m)$ ?

1 Answer

Explanation:

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How do you multiply (1m2−m)+(1m)=5m2−m(1/(m^2-m)) + (1/m)=5/(m^2-m)?

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How do you multiply $(\frac{1}{m^{2} - m}) + (\frac{1}{m}) = \frac{5}{m^{2} - m}$ $(1/(m^2-m)) + (1/m)=5/(m^2-m)$ ?