Solve: The quantity 2 x minus 20 divided by 3 = 2x

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Solve: The quantity 2 x minus 20 divided by 3 = 2x

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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(2x-20)/3 = 2x start by multiplying both sides by 3
by both sides do you mean .. both sides of the equation ?
yes

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Other answers:

\[\large \frac{2x-20}{3} =2x \] yes multiply 3 throughout the equation
\(\bf {\color{brown}{ \cancel{3}}}\cdot \cfrac{2x-20}{\cancel{3}} ={\color{brown}{ 3}}\cdot 2x\)
so would i have 6x-60=6x ?
by multiplying 3 on both sides... the 3 for the left side ends up being canceled out
so .... 2x-20= 3 times 2x ? im lost .. im not good at math
or another way of seeing that there is cancellation \[\large \frac{(3)(2x-20)}{3} =2x (3)\] \[\large \frac{(6x-60)}{3} =2x (3)\] \[\large 2x-20=6x\]
yeah you will have 2x-20= 3(2x) which becomes 2x-20=6x
okay i understand that part now
now we just have to have all variables to the left and all numbers to the right so add 20 to both sides and subtract 6x to both sides
can i solve it like a regular equation ....|dw:1438127165206:dw|
sorry its tso sloppy
so*
it's ok. I just want to check... just to be sure
|dw:1438127376054:dw| it's correct!
i think thats right its in the answer box which is −5 −3 2 13
the answer should be the same even if the terms are rearranged (variables to the left, numbers to the right) |dw:1438127444700:dw|
thank youu soo much you are a really good helper :)
^_^ you're welcome

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