Solve 2(x + 1) = 2x + 2.

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Solve 2(x + 1) = 2x + 2.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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is the answer no solution? options: all real numbers no solution 0 1
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since after you distribute the 2 on the left side, both side are the same so it is all real numbers or infinitely many solutions

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\[2x+2=2x+2\]
all values of x are solutions here.
Hope this helped!
but when you subtract the two together, are they not equal to eachother I got 2x=2x at the end
yes, 2x=2x is a true statement. they are equal
so i would not be no solution because they cancel out?
exactly, if you subtract 2x from each side it would be 0=0 which both sides are equal. if for instance x=5 than when you substitute in 5 for x, it comes out to 10 on each side. therefore both sides are equal to eachother.
and if you divided each side by 2 which is what you would/should do than it would be x=x where x could equal any number!
yea thats what i did so it would be no solution
no, if x=x than x could be any number so it is all real numbers/infinitely many solutions!
all real numbers then
yes
sorry if my examples of 0=0 or 10=10 mixed you up. i was just showing that both sides were equal so there had to be a solution
alright thanks that helped
no problem!
just know in the future, when you get a question like this where both side are the some it is all real numbers or infinitely many solutions. don't question it!
ok thanks

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