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

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

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

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

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

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

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

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

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

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

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

gangescom Group TitleBest ResponseYou've already chosen the best response.0
dw:1347312457904:dw
 2 years 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)
 2 years ago

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

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