isabeli14 isabeli14
  • 01-12-2020
  • Mathematics
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2b-3/b-2 = 3b/b+2 ... Solutions by cross multiplying

ANSWER > 1, 6
- Neither 1 or 6 are extraneous

Respuesta :

minn20 minn20
  • 01-12-2020

Step-by-step explanation:

(2b-3)/(b-2)(b+2) = 3b(b-2)/(b+2)(b-2)

2b²+4b-3b-6/b²-4=3b²-6b/b²-4

2b²+b- 6= 3b²-6b

move square to square unknown to unknown

-b²+7b-6/b²-4

i also dk this is true or false

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