Verified answer

4(x − 1) − (x + 2) = 12

4x-4-x-2=12

3x-6=12

3x=12+6

x=18/3

x=6

d.6

6

4(x − 1) − (x + 2) = 12

4x - 4 -x -2 = 12

3x - 6 = 12

3x = 18

x = 6

4(x-1) - (x+2) = 12

4x - 4 - x - 2 = 12

3x - 6 = 12

3x = 18

x = 6

4x - 4 - x - 2 = 12

3x - 6 = 12

3x = 18

x = 6

hmmm...likely...besides the indisputable fact that the only which is going outstanding for the wizard or the only he's exceptional proper for typically is the only chosen. it isn't going besides the indisputable fact that that the point of compatibility between both wands will be realtively severe! One is absolute to be extra properly matched. also, each wand has a center, and a particular lentgh. The length of a wand relies upon on the length of the wizards arm, and the middle finally relies upon on what style of magic the wizard prefers. If a wizard is extra apt for Transfiuration , as an celebration, then that individual will be extra ideal to a particular center. and also of course if the wand is gained, then that wand will supply you with its allegiance...making you an proprietor of two wands...very similar to Harry and Dumbledore! So i imagine its feasible, yet really unlikely! And it very not often occurs (i imagine)

4x-4-x-2=12

3x=18

x=6

4(x-1)-(x+2)=12

4x-4 - (x+2)=12

+4 +4

------------------------

4x - (x+6) =12

-x -x

------------------------

3x - 6 = 12

+6 +6

-------------------------

3x = 18

18/3 = 6

so, x=6

14/3

I think