How do you solve #4(5b +2)-b=2+8(3b-8)#?

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Answer 1 · 2017-06-29T05:07:40

#b=14#

Both brackets neednto be expanded first, where #a(b+c)-=ab+ac#

This means that #4(5b+2)-=(4*5b)+(4*2)=20b+8#

It also means that #8(3b-8)-=(8*3b)+(8*-8)=24b-64#

You now have #20b+8-b=2+24b-64#. Put all b's on the left hand side and numbers on the right.

#20b+8-b=2+24b-64#
#20b-b-24b=-8+2-64#
#-5b=-70#

Now divide both sides by #-5#.

#(-5b)/(-5)=(-70)/(-5)#
#b=14#

Answer 2 · 2017-06-29T05:12:14

#b=14#

#4(5b+2)-b=2+8(3b-8)#

First open the brackets and simplify by multiplying each term with the number outside the bracket.

#20b+8-b=2+24b-64#

Simplify each side.

#19b+8=24b-62#

Add #62# to each side.

#19b+70=24b#

Subtract #19b# from each side.

#70=5b#

Divide both sides by #5#.

#14=b#