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Mathematics 7 Online
OpenStudy (anonymous):

4/5(2x-1)>12

OpenStudy (anonymous):

Are you solving for x?

OpenStudy (anonymous):

Excuse me, is it (4/5)*(2x-1)>12 or 4/[5(2x-1)]>12? could you make that clearer

OpenStudy (anonymous):

if it is (4/5)*(2x-1)>12 then x>8; if it is 4/[5(2x-1)]>12 then x>(4/5)

OpenStudy (anonymous):

Yes, solving for x, and the first of the options is what it is supposed to be. :Thank you. :)

OpenStudy (anonymous):

It told me the answer was incorrect. :( Here is a similar equation, with as far as I can get on my own: 2/3(3x-6)>12 6/3x-12/3>12 12(6/3x-12/3)>12(12) (72/3)x-(144/3)>144 I can not seem to get to the answer from here. I'm sure it's probably easy, but I need it explained in simple terms.

OpenStudy (anonymous):

2/3(3x-6)>12 3x-6>18 3x>24 x>8

OpenStudy (anonymous):

Also needing assistance with compound inequalities. Will be posting one soon. I don't just want an answer. I want someone to teach me (simply) how to solve them, step-by-step. Note: The compound inequalities have some odd fractions, so I may have to explain them differently.

OpenStudy (anonymous):

Dhat: Where did the 2/3 go?

OpenStudy (anonymous):

I multiplied by 3/2 on both sides.

OpenStudy (anonymous):

OK, I see. Thank you. :)

OpenStudy (anonymous):

you are welcome

OpenStudy (anonymous):

Would you like me to post the compund inequality here, or as a new question?

OpenStudy (anonymous):

new thread please

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