Solve 2[1-1/3(x-1+x/3)]=1/2(2x-10-3x/3)quad(8)frac{1}{2}

$$2 [ 1 - \frac { 1 } { 3 } ( x - \frac { 1 + x } { 3 } ) ] = \frac { 1 } { 2 } ( 2 x - \frac { 10 - 3 x } { 3 } ) \quad ( 8 ) \frac { 1 } { 2 }$$

Solve for x

$$x = \frac{40}{29} = 1\frac{11}{29} \approx 1.379310345$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=3\left(2x-\frac{10-3x}{3}\right)\times 8\times \frac{1}{2}$$

Multiply 3 and 8 to get 24.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=24\left(2x-\frac{10-3x}{3}\right)\times \frac{1}{2}$$

Multiply 24 and \frac{1}{2} to get \frac{24}{2}.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=\frac{24}{2}\left(2x-\frac{10-3x}{3}\right)$$

Divide 24 by 2 to get 12.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=12\left(2x-\frac{10-3x}{3}\right)$$

Use the distributive property to multiply 12 by 2x-\frac{10-3x}{3}.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=24x+12\left(-\frac{10-3x}{3}\right)$$

Cancel out 3, the greatest common factor in 12 and 3.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=24x-4\left(10-3x\right)$$

Use the distributive property to multiply -4 by 10-3x.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=24x-40+12x$$

Combine 24x and 12x to get 36x.

$$12\left(1-\frac{1}{3}\left(x-\frac{1+x}{3}\right)\right)=36x-40$$

Divide each term of 1+x by 3 to get \frac{1}{3}+\frac{1}{3}x.

$$12\left(1-\frac{1}{3}\left(x-\left(\frac{1}{3}+\frac{1}{3}x\right)\right)\right)=36x-40$$

To find the opposite of \frac{1}{3}+\frac{1}{3}x, find the opposite of each term.

$$12\left(1-\frac{1}{3}\left(x-\frac{1}{3}-\frac{1}{3}x\right)\right)=36x-40$$

Combine x and -\frac{1}{3}x to get \frac{2}{3}x.

$$12\left(1-\frac{1}{3}\left(\frac{2}{3}x-\frac{1}{3}\right)\right)=36x-40$$

Use the distributive property to multiply -\frac{1}{3} by \frac{2}{3}x-\frac{1}{3}.

$$12\left(1-\frac{1}{3}\times \frac{2}{3}x-\frac{1}{3}\left(-\frac{1}{3}\right)\right)=36x-40$$

Multiply -\frac{1}{3} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$12\left(1+\frac{-2}{3\times 3}x-\frac{1}{3}\left(-\frac{1}{3}\right)\right)=36x-40$$

Do the multiplications in the fraction \frac{-2}{3\times 3}.

$$12\left(1+\frac{-2}{9}x-\frac{1}{3}\left(-\frac{1}{3}\right)\right)=36x-40$$

Fraction \frac{-2}{9} can be rewritten as -\frac{2}{9} by extracting the negative sign.

$$12\left(1-\frac{2}{9}x-\frac{1}{3}\left(-\frac{1}{3}\right)\right)=36x-40$$

Multiply -\frac{1}{3} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$12\left(1-\frac{2}{9}x+\frac{-\left(-1\right)}{3\times 3}\right)=36x-40$$

Do the multiplications in the fraction \frac{-\left(-1\right)}{3\times 3}.

$$12\left(1-\frac{2}{9}x+\frac{1}{9}\right)=36x-40$$

Convert 1 to fraction \frac{9}{9}.

$$12\left(\frac{9}{9}-\frac{2}{9}x+\frac{1}{9}\right)=36x-40$$

Since \frac{9}{9} and \frac{1}{9} have the same denominator, add them by adding their numerators.

$$12\left(\frac{9+1}{9}-\frac{2}{9}x\right)=36x-40$$

Add 9 and 1 to get 10.

$$12\left(\frac{10}{9}-\frac{2}{9}x\right)=36x-40$$

Use the distributive property to multiply 12 by \frac{10}{9}-\frac{2}{9}x.

$$12\times \frac{10}{9}+12\left(-\frac{2}{9}\right)x=36x-40$$

Express 12\times \frac{10}{9} as a single fraction.

$$\frac{12\times 10}{9}+12\left(-\frac{2}{9}\right)x=36x-40$$

Multiply 12 and 10 to get 120.

$$\frac{120}{9}+12\left(-\frac{2}{9}\right)x=36x-40$$

Reduce the fraction \frac{120}{9} to lowest terms by extracting and canceling out 3.

$$\frac{40}{3}+12\left(-\frac{2}{9}\right)x=36x-40$$

Express 12\left(-\frac{2}{9}\right) as a single fraction.

$$\frac{40}{3}+\frac{12\left(-2\right)}{9}x=36x-40$$

Multiply 12 and -2 to get -24.

$$\frac{40}{3}+\frac{-24}{9}x=36x-40$$

Reduce the fraction \frac{-24}{9} to lowest terms by extracting and canceling out 3.

$$\frac{40}{3}-\frac{8}{3}x=36x-40$$

Subtract 36x from both sides.

$$\frac{40}{3}-\frac{8}{3}x-36x=-40$$

Combine -\frac{8}{3}x and -36x to get -\frac{116}{3}x.

$$\frac{40}{3}-\frac{116}{3}x=-40$$

Subtract \frac{40}{3} from both sides.

$$-\frac{116}{3}x=-40-\frac{40}{3}$$

Convert -40 to fraction -\frac{120}{3}.

$$-\frac{116}{3}x=-\frac{120}{3}-\frac{40}{3}$$

Since -\frac{120}{3} and \frac{40}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{116}{3}x=\frac{-120-40}{3}$$

Subtract 40 from -120 to get -160.

$$-\frac{116}{3}x=-\frac{160}{3}$$

Multiply both sides by -\frac{3}{116}, the reciprocal of -\frac{116}{3}.

$$x=-\frac{160}{3}\left(-\frac{3}{116}\right)$$

Multiply -\frac{160}{3} times -\frac{3}{116} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{-160\left(-3\right)}{3\times 116}$$

Do the multiplications in the fraction \frac{-160\left(-3\right)}{3\times 116}.

$$x=\frac{480}{348}$$

Reduce the fraction \frac{480}{348} to lowest terms by extracting and canceling out 12.

$$x=\frac{40}{29}$$