What is the answer to 3+2[1+2(2x-2)]=6(x+3) ?

The answer to 3+2[1+2(2x-2)]=6(x+3) is x= 21/2

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3+2[1+2(2x-2)]=6(x+3)

3 + 2 [1 + 2 (2 x - 2)] = 6 (x + 3)

x = \frac{21}{2}

Decimal Notation

x = 10.5

Solution steps

3 + 2 [1 + 2 (2 x - 2)] = 6 (x + 3)

Expand

3 + 2 (1 + 2 (2 x - 2)) : 8 x - 3

Expand

6 (x + 3) : 6 x + 18

8 x - 3 = 6 x + 18

Move

3

to the right side

8 x = 6 x + 21

Move

6 x

to the left side

2 x = 21

Divide both sides by

2

x = \frac{21}{2}

r^{2} - s^{2}

60 \cdot 1 0^{- 6} = 30 \cdot 1 0^{6} (x - 0.4)^{2}

- 98

140000000 = - 66.67 x^{2} + 3133.33 x - 26000

y^{3} - y

What is the answer to 3+2[1+2(2x-2)]=6(x+3) ?
The answer to 3+2[1+2(2x-2)]=6(x+3) is x= 21/2
Solve for x: 3+2[1+2(2x-2)]=6(x+3)
The solution is x= 21/2