How do you solve $\frac{1}{r} - \frac{6}{5 r + 2} = 1$ $\frac { 1} { r } - \frac { 6} { 5r + 2} = 1$ ?

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How do you solve $\frac{1}{r} - \frac{6}{5 r + 2} = 1$ $\frac { 1} { r } - \frac { 6} { 5r + 2} = 1$ ?

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Answer 1 · 2018-03-24T05:31:50

$r = - \frac{3}{10} \pm \frac{1}{5} i$ $r=-3/10+-1/5i$ , Or if those Don't work, There are No Solutions!

Explanation:

First, Multiply By the denominators.
witch Are,
$r and 5 r + 2$ $r and 5r+2$
so, The equation Would be,

$5 r + 2 - 6 r = 5 r^{2} + 2 r$ $5r+2-6r=5r^2+2r$

Re-aranging, We Get

$- 5 r^{2} - 3 r + 2 = 0$ $-5r^2-3r+2=0$

And by the quadratic Formula, The solution Is,

$r = \frac{3 \pm \sqrt{3^{2} - 15}}{- 10} = \frac{3 \pm 2 i}{- 10} = - \frac{3}{10} \pm \frac{1}{5} i$ $r=(3+-sqrt(3^2-15))/-10=(3+-2i)/-10=-3/10+-1/5i$

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How do you solve $\frac{1}{r} - \frac{6}{5 r + 2} = 1$ $\frac { 1} { r } - \frac { 6} { 5r + 2} = 1$ ?

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Explanation:

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How do you solve $\frac{1}{r} - \frac{6}{5 r + 2} = 1$ $\frac { 1} { r } - \frac { 6} { 5r + 2} = 1$ ?