What is the answer to -[2(3x-3)-(1+x)]-[5(3-2x)-(1+x)]=0 ?

The answer to -[2(3x-3)-(1+x)]-[5(3-2x)-(1+x)]=0 is x= 7/6

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-[2(3x-3)-(1+x)]-[5(3-2x)-(1+x)]=0

- [2 (3 x - 3) - (1 + x)] - [5 (3 - 2 x) - (1 + x)] = 0

x = \frac{7}{6}

Decimal Notation

x = 1.16666 \dots

Solution steps

- [2 (3 x - 3) - (1 + x)] - [5 (3 - 2 x) - (1 + x)] = 0

Expand

- (2 (3 x - 3) - (1 + x)) - (5 (3 - 2 x) - (1 + x)) : 6 x - 7

6 x - 7 = 0

Move

7

to the right side

6 x = 7

Divide both sides by

6

x = \frac{7}{6}

12 b = 24

\frac{n}{5} = 9

2 (x - 3) = 6

2 (x - 3) = 0

- (x + 1) - 2 x = x - (12 + (3 - x))

What is the answer to -[2(3x-3)-(1+x)]-[5(3-2x)-(1+x)]=0 ?
The answer to -[2(3x-3)-(1+x)]-[5(3-2x)-(1+x)]=0 is x= 7/6
Solve for x: -[2(3x-3)-(1+x)]-[5(3-2x)-(1+x)]=0
The solution is x= 7/6