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

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

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

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

x = 1

Solution steps

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

Simplify

1 + 3 x - 2 (x + 2) + 3 x : 6 x - 2 (x + 2) + 1

1 - 2 (6 x - 2 (x + 2) + 1) = - 1

Move

1

to the right side

- 2 (6 x - 2 (x + 2) + 1) = - 2

Divide both sides by

- 2

4 x - 3 = 1

Move

3

to the right side

4 x = 4

Divide both sides by

4

x = 1

(4^{3})^{\frac{1}{2}}

7.4 \cdot (\frac{1.48 + 1.13}{6})

t (t - 1) = 0

7 + a < 11

(\frac{4}{5})^{- 3}

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