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Tombrokaw
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How do I solve this problem 7(the lesser than or equal to sign) 2x+3(<) 5
 one year ago
 one year ago
Tombrokaw Group Title
How do I solve this problem 7(the lesser than or equal to sign) 2x+3(<) 5
 one year ago
 one year ago

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gangescom Group TitleBest ResponseYou've already chosen the best response.0
you are going to need to set up two equations
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
one 7(less than or equal sign) 2X+3 two 2X+3(<)5 you will get two answers it would look like this for example 6<x>9
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
Alright Thank you, I will tell you what I came up with so far
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
for my solutions I have(5,infinity symbol] and (5/2,infinity symbol)
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
Infinity?
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
ok lets work Equation one 7(Less than or equal) 2X+3
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
subtract 3 from both sides
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
I end up with 2 is less than or equla to x
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
you should get 10 since you subtracted 3 and you have a 7 on the other side you get 10
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
7 (less than equal) 2X+3 3 3 =10 (less than equal) 2x
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
then divide by 2 and we get 5
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
yup good job!
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
do the same for the other equation
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
but I am supposed to express each in interval notation
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
and for the other equation x<5/2
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
you should get 1 for the other equation
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
2X+3<5 3<3 2X<2 divide by 2 on both side 2 2 X=1
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
dw:1347312457904:dw
 one year ago

gangescom Group TitleBest ResponseYou've already chosen the best response.0
your interval notation sould look something like this its a bad drawing, and than draw the arrows (left or right)
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
The solution would then look x<1 for the other one
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
if you started with 7 ≤ 2x+3 < 5 you should get, after simplifying, 5 ≤ x < 1 in interval notation, this is [5,1) the square brackets indicates 5 is part of the interval. ) means the upper value is a limit, approached but not reached.
 one year ago

Tombrokaw Group TitleBest ResponseYou've already chosen the best response.0
Alright, thank you
 one year ago
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