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

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Answer 1 · 2018-03-24T05:31:50

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

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

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

Re-aranging, We Get

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

And by the quadratic Formula, The solution Is,

$r=(3+-sqrt(3^2-15))/-10=(3+-2i)/-10=-3/10+-1/5i$

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