How do you solve #-2( 6+ s ) \geq - 15- 23#?

1 Answer
smendyka
Dec 7, 2016

#s <= 13#

Explanation:

First, combine like terms and expand the term in parenthesis:

#-2*6 + -2s >= -38#

#-12 - 2s >= -38#

Next, isolate the #s# term:

#-12 + 12 - 2s >= -38 + 12#

#0 - 2s >= -26#

#-2s >= -26#

#Now we can solve for #s#. However, we must remember, when dealing with inequalities if you multiply or divide by a negative number it reverses the inequality:

#(-2s)/-2 <= (-26)/(-2)#

#(cancel(-2)s)/cancel(-2) <= 13#

#s <= 13#

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