How do you solve #9-2x \le 3 or 3x+10 \le 6-x#?

1 Answer
May 27, 2018

Answer:

See explanation

Explanation:

We have two conditions that are combined to define the limit of values that may be assigned to #x#

Condition 1: #9-2x<=3#
Condition 2: #2x+10<=6-x#

Consider condition 1

Add #2x# to both sides

#9color(white)("dd")ubrace(-2x+2x)<=3+2x#

#9 color(white)("ddd")+ 0color(white)("dddd")<=2x+3#

Subtract 3 from both sides

#9-3<=2x + 0#

Divide both sides by 2

#(9-3)/2<=2/2color(white)(.)x#

#3<=x#
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Consider condition 2

Add #x# to both sides

#3x+10<=6#

Subtract 10 from both sides

#3x<=-4#

Divide both sides by 3

#x<=-4/3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Combining these we have:

#-4/3 >= x>=3#

In other words #x# does not take on any values between and excluding #-4/3 and 3#

Tony B