1 answer

How do you solve #1/3x + 2 >3# and #-3x > 12#?

Question:

How do you solve #1/3x + 2 >3# and #-3x > 12#?

Answers

# 3 < x <4#

Explanation:

Treat inequalities the same as equations, except if you multiply or divide by a negative value, in which case the inequality sign changes around.

#1/3x+2 >3color(white)(xxxxxxxxxxxxxx)-3x >12#

#1/3x > 1" "larr xx 3color(white)(xxxxxxxxxxx) 12>3x" "larr div3#

#x > 3color(white)(xxxxxxxxxxxxxxxxxxxxxx)4 >x#

Combine the two inequalities,

#3 < x and x < 4#

#:. 3 < x <4#

.

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