How do you solve #(b+6)/(4b^2)+3/(2b^2)=(b+4)/(2b^2)# and check for extraneous solutions?

1 Answer
Dec 18, 2016

Answer:

b = 4

Explanation:

#(b+6)/(4b^2)+3/(2b^2)=(b+4)/(2b^2)#
multiply both sides by #4b^2#
#4b^2*(b+6)/(4b^2)+4b^2*3/(2b^2)=4b^2(b+4)/(2b^2)#
#b+6+6=2b+8#
#b+12=2b+8#
#b-2b=8-12#
#-b=-4#
#b=4#
check:
substitute b=4
#((4)+6)/(4(4)^2)+3*(4(4^2))/(2(4^2)#=#(4+4)/(2(4^2)#
#10/64+3/32=8/32#
#5/32+3/32=8/32#
#8/32=8/32#