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

2 Answers
Sep 6, 2017

#s=-34#

Explanation:

#\color(steelblue)(4(s+8))-3s=-2#
#\color(steelblue)(4(s)+4(8))-3s=-2# Distributive property.
#\color(steelblue)(4s+32)-3s=-2# Simplify, then combine like terms.

#\color(indianred)(4s)\color(seagreen)(+32)\color(indianred)(-3s)=\color(seagreen)(-2)# Identify like terms. Start with the coefficient variable terms.
#\color(indianred)(4s-3s)\color(seagreen)(+32)=\color(seagreen)(-2)# Combine previously mentioned terms.
#\color(indianred)(s)\color(seagreen)(+32)=\color(seagreen)(-2)# Simplify, then do the same for the non-variable terms.

#s=\color(seagreen)(-2-32)# Move all non-variable terms to the right.
#s=\color(seagreen)(-34)# Combine those terms. Now you have your answer:

#s=-34#

Sep 6, 2017

#=>s=-34#

Explanation:

#4(s+8)-3s=-2#
#=>4s+32-3s=-2#
#=>s+32=-2#

Let's add #-32# to both side:
#s=-2-32#
#=>s=-34#