How do you solve #4(l-1)+2(l+1)=0#?

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Answer 1 · 2015-06-30T19:37:50

Distribute the #4# and #2#, group like terms, then solve for #l#.

#4(l-1)+2(l+1)=0#

Distribute the #4# and #2#.

#4l-4+2l+2=0#

Group like terms.

#4l+2l-4+2=0#

#6l-2=0#

Add #2# to both sides.

#6l=2#

Divide both sides by #6#.

#l=2/6#

Simplify.

#l=1/3#