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I have solved the equation 2/3-2r=1/3r+7/3 to -3/7 is this correct?

 Aug 21, 2016
 #1
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Solve for r:
2/3-2 r = r/3+7/3

 

Put each term in 2/3-2 r over the common denominator 3: 2/3-2 r = 2/3-(6 r)/3:
2/3-(6 r)/3 = r/3+7/3

 

2/3-(6 r)/3 = (2-6 r)/3:
(2-6 r)/3 = r/3+7/3

 

r/3+7/3 = (r+7)/3:
(2-6 r)/3 = (r+7)/3

 

Multiply both sides by 3:
2-6 r = r+7

 

Subtract r from both sides:
2+(-6 r-r) = (r-r)+7

 

-6 r-r = -7 r:
-7 r+2 = (r-r)+7

 

r-r = 0:
2-7 r = 7

 

Subtract 2 from both sides:
(2-2)-7 r = 7-2

 

2-2 = 0:
-7 r = 7-2

 

7-2 = 5:
-7 r = 5

 

Divide both sides of -7 r = 5 by -7:
(-7 r)/(-7) = 5/(-7)

 

(-7)/(-7) = 1:
r = 5/(-7)

 

Multiply numerator and denominator of 5/(-7) by -1:
Answer: |r = (-5)/7

 Aug 21, 2016
 #2
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 2/3-2r=1/3r+7/3, solve for r

 

-2r - 1/3r =7/3 - 2/3

-2 1/3r =5/3

-7/3r =5/3 multiply both sides by - 3/7

r =5/3 x - 3/7

r= - 15/21

r= - 5/7 

 Aug 21, 2016

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