What is the answer to (3x-4)^2+2-(1-x)=(1-3x)^2 ?

The answer to (3x-4)^2+2-(1-x)=(1-3x)^2 is x= 16/17

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(3x-4)^2+2-(1-x)=(1-3x)^2

(3 x - 4)^{2} + 2 - (1 - x) = (1 - 3 x)^{2}

x = \frac{16}{17}

Decimal Notation

x = 0.94117 \dots

Solution steps

(3 x - 4)^{2} + 2 - (1 - x) = (1 - 3 x)^{2}

Expand

(3 x - 4)^{2} + 2 - (1 - x) : 9 x^{2} - 23 x + 17

Expand

(1 - 3 x)^{2} : 1 - 6 x + 9 x^{2}

9 x^{2} - 23 x + 17 = 1 - 6 x + 9 x^{2}

Move

17

to the right side

9 x^{2} - 23 x = 9 x^{2} - 6 x - 16

Subtract

9 x^{2} - 6 x

from both sides

9 x^{2} - 23 x - (9 x^{2} - 6 x) = 9 x^{2} - 6 x - 16 - (9 x^{2} - 6 x)

Simplify

- 17 x = - 16

Divide both sides by

- 17

x = \frac{16}{17}

x, 3 x + 6 y = - 12

(19) (12) = (10 + 9) (10 + 2)

x - 9 = - 20

x - 9 = - 13

x, 5 x + 2 y = λ x

What is the answer to (3x-4)^2+2-(1-x)=(1-3x)^2 ?
The answer to (3x-4)^2+2-(1-x)=(1-3x)^2 is x= 16/17
Solve for x: (3x-4)^2+2-(1-x)=(1-3x)^2
The solution is x= 16/17