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7±[2(x+1)-3(x-2/3)]

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2 Answers

I am working under the premise that x-2/3 does NOT mean (x-2)/3...

7 +- [2(x+1) - 3(x-2/3)]

7 +- (2x + 2 - 3x + 2)

7 +- (2x -3x + 2 + 2)

7 +- (4-x)

Two possibilities:
7 + (4-x) = 11 - x
7 - (4-x) = 7 -4 + x = 3 +x


11-x and 3+x
You have to work from the innermost parenthesis first, so you have to distribute the '3' to the contents of its parenthesis by multiplication: 3(x-2/3) = 3x - 2
 
then do the same with 2(x+1): 2x + 2
 
Substitute the distributed values above into their spots in their original equations:
 
7 ± [2x+2-(3x-2)]  or  7 ± [2x + 2 - (3x - 2)] = 7 ± [2x +2 - 3x + 2]
 
Solve for the part enclosed by parenthesis (re-write): 2x-3x +2 +2 = -x + 4
 
Substitute that answer again: 7 ±[ -x + 4] and get the answers:
 
       7+(-x+4) = -x+11 
                 and
       7-(-x+4) = 7+x-4 = x + 3