How do you solve and graph #-2a+3>=6a-1>3a-10#?

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Tony B Share
Feb 24, 2018

Answer:

#-3 < a <= 1/2#

Explanation:

#color(blue)("Consider: "-2a+3>=6a-1)#

Add #2a# to both sides

#3>=8a-1#

Add 1 to both sides

#4>=8a#

Divide both sides by 8

#1/2>=a#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider: "6a-1>3a-10)#

Subtract #3a# from both sides

#3a-1> - 10#

Add 1 to both sides

#3a > -9#

Divide both sides by 3

#a > -3#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("putting it all together")#

In the same order as in the question.

#1/2>=a > -3#

Lets 'rotate' it horizontally #180^o#

#color(red)("THE SOLUTION "-> bar(ul(|color(white)("d")-3 < a <= 1/2color(white)("d") |)))#

You use a 'hollow' circle at the end for 'greater than' or 'less than'.

You use a 'filled in' circle at the end for 'less than or equal to' or 'greater than or equal to'.

Tony B
#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

#color(blue)("IF YOU WERE TO PLOT THESE AND ASSIGN THE VALUES TO y")#

The feasible region for #y# is the shaded area

To construct this I used:
#ubrace(-2a+3>=6a-1>3a-10)#

#-2x+3>=y>3x-10#

AND

#ubrace(-2a+3>=6a-1>3a-10)#
#color(white)("dddddd")y>=6x-1#

#y<=-2x+3#
#y>=color(white)(-)6x-1#
#y>color(white)(-)3x-10#

#x>=-3 larr" Needed to control the lower bound of "a#

Tony B

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Write your answer here...
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Answer

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Feb 24, 2018

Answer:

Answer is #-3 < a<=1/2#

Explanation:

#-2a+3>=6a-1>3a-10#

means #-2a+3>=6a-1#

i.e. #3+1>=6a+2a#

or #8a<=4# i.e. #a<=1/2#

Also #6a-1>3a-10#

or #6a-3a> -10+1#

or #3a> -9# i.e. #a> -3#

Hence answer is #-3<a<=1/2#

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