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Mathematics 19 Online
OpenStudy (anonymous):

Solve |2y - 2| = y. A. {2/3, 2} B. {2, 3} C. {2, 5} D. {1/3, 2} Solve |2y - 2| = y. A. {2/3, 2} B. {2, 3} C. {2, 5} D. {1/3, 2} @Mathematics

OpenStudy (anonymous):

2y-2=y and 2y-2=-y 2y=y+2 and 2y=2-y y=2 and y=2/3

OpenStudy (anonymous):

so A

OpenStudy (anonymous):

Okay thank you

OpenStudy (anonymous):

do you understand how I got that?

OpenStudy (anonymous):

No not really

OpenStudy (anonymous):

ok well this is how you do it abs(whatever)= something this means that (whatever)=soemthign and (whatever)=negative something so forexample abs(x)=3 means x=3 and x=-3 also if it said abs(x) + 2=3, this means you have to isolate the absolute to just itself so it would be abs(x)=1 [subtract the 2] Does that help?

OpenStudy (anonymous):

Um yea actually it does.

OpenStudy (anonymous):

Okay so do you think you can help me with the question i just posted

OpenStudy (anonymous):

sure

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