How do you solve $- \frac{3}{2 g + 1} = \frac{3}{g + 6}$ $-3/(2g+1)=3/(g+6)$ ?

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Answer 1 · 2018-05-25T10:12:54

$- \frac{21}{9}$ $-21/9$

$- \frac{3}{2 g + 1} = \frac{3}{g + 6}$ $-3/(2g+1)=3/(g+6)$

or, $- 3 (g + 6) = 3 (2 g + 1)$ $-3(g+6)=3(2g+1)$

or, $- 3 g - 18 = 6 g + 3$ $-3g-18=6g+3$

or, $- 3 g - 6 g = 3 + 18$ $-3g-6g=3+18$

or, $- 9 g = 21$ $-9g=21$

or, $g = - \frac{21}{9} \approx - 2.33$ $g=-21/9approx-2.33$

How do you solve −32g+1=3g+6-3/(2g+1)=3/(g+6)?