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Mathematics 23 Online
OpenStudy (anonymous):

2x + 1 < 1 OR x + 5 > 8

jimthompson5910 (jim_thompson5910):

2x + 1 < 1 OR x + 5 > 8 2x < 1 - 1 OR x > 8 - 5 2x < 0 OR x > 3 x < 0/2 OR x > 3 x < 0 OR x > 3

OpenStudy (anonymous):

how did you do that? you switched the signs?

jimthompson5910 (jim_thompson5910):

which step?

OpenStudy (anonymous):

2nd

jimthompson5910 (jim_thompson5910):

the signs are the same as the first step

OpenStudy (anonymous):

no?

jimthompson5910 (jim_thompson5910):

I'm dealing with each inequality individually

jimthompson5910 (jim_thompson5910):

I solved 2x + 1 < 1 to get x < 0

jimthompson5910 (jim_thompson5910):

separately, I solved x + 5 > 8 to get x > 3

OpenStudy (anonymous):

but if you do it to one side dont you do it to the others?

jimthompson5910 (jim_thompson5910):

that's what I'm doing when I did 2x < 1-1, I subtracted 1 from both sides (first inequality only)

OpenStudy (anonymous):

oh.

jimthompson5910 (jim_thompson5910):

going from 2x + 1 < 1 to 2x < 1-1, the sign does not change

OpenStudy (anonymous):

thanks

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