How do you solve #x + 3/2 = 8/x -3#?

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Answer 1 · 2017-10-12T02:36:36

# 2x^2 +9x -16 =0#

#x + 3/2 = 8/x -3#

Make denominators equal

#x(2/2) + 3/2 = 8/x -3(x/x)#

#(2x + 3)/2 = (8-3x)/x #

Transposing denominators:

#x(2x+3) = 2(8-3x)#

# 2x^2 +3x = 16 - 6x#

# 2x^2 +9x -16 =0#