How do you solve $\frac{3}{n} + \frac{3}{n^{2} - 5 n} = \frac{1}{n^{2} - 5 n}$ $\frac { 3} { n } + \frac { 3} { n ^ { 2} - 5n } = \frac { 1} { n ^ { 2} - 5n }$ ?

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Answer 1 · 2017-08-14T15:09:30

$Answer : n = \frac{13}{3}$ $"Answer :"n=13/3$

$\frac{3}{n} + \frac{3}{n^{2} - 5 n} = \frac{1}{n^{2} - 5 n}$ $3/n+3/(n^2-5 n)=1/(n^2-5 n)$

$\frac{3}{n} = \frac{1}{n^{2} - 5 n} - \frac{3}{n^{2} - 5 n}$ $3/n=1/(n^2-5 n)-3/(n^2-5 n)$

$\frac{3}{n} = \frac{- 2}{n^{2} - 5 n}$ $3/n=(-2)/(n^2-5 n)$

$- 2 n = 3 (n^{2} - 5 n)$ $-2 n=3(n^2-5 n)$

$- 2 n = 3 n^{2} - 15 n$ $-2 n=3 n^2-15n$

$3 n^{2} - 15 n + 2 n = 0$ $3n^2-15n+2n=0$

$3 n^{2} - 13 n = 0$ $3 n^2-13n=0$

$3 n^{2} = 13 n$ $3n^2=13n$

$3cancel(n^2)=13cancel(n)$

$3n=13$

$(cancel(3)n)/cancel(3)=(13)/3$

$n=13/3$

How do you solve \frac { 3} { n } + \frac { 3} { n ^ { 2} - 5n } = \frac { 1} { n ^ { 2} - 5n }3n+3n2−5n=1n2−5n\frac { 3} { n } + \frac { 3} { n ^ { 2} - 5n } = \frac { 1} { n ^ { 2} - 5n }?