Solve 4/3-10x+4-[2/3(x-4)+2x+1/3]=-5x+2/3(x-1

$$\frac { 4 } { 3 } - 10 x + 4 - [ \frac { 2 } { 3 } ( x - 4 ) + 2 x + \frac { 1 } { 3 } ] = - 5 x + \frac { 2 } { 3 } ( x - 1$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Convert 4 to fraction \frac{12}{3}.

$$\frac{4}{3}-10x+\frac{12}{3}-\left(\frac{2}{3}\left(x-4\right)+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Since \frac{4}{3} and \frac{12}{3} have the same denominator, add them by adding their numerators.

$$\frac{4+12}{3}-10x-\left(\frac{2}{3}\left(x-4\right)+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Add 4 and 12 to get 16.

$$\frac{16}{3}-10x-\left(\frac{2}{3}\left(x-4\right)+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Use the distributive property to multiply \frac{2}{3} by x-4.

$$\frac{16}{3}-10x-\left(\frac{2}{3}x+\frac{2}{3}\left(-4\right)+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Express \frac{2}{3}\left(-4\right) as a single fraction.

$$\frac{16}{3}-10x-\left(\frac{2}{3}x+\frac{2\left(-4\right)}{3}+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Multiply 2 and -4 to get -8.

$$\frac{16}{3}-10x-\left(\frac{2}{3}x+\frac{-8}{3}+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Fraction \frac{-8}{3} can be rewritten as -\frac{8}{3} by extracting the negative sign.

$$\frac{16}{3}-10x-\left(\frac{2}{3}x-\frac{8}{3}+2x+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Combine \frac{2}{3}x and 2x to get \frac{8}{3}x.

$$\frac{16}{3}-10x-\left(\frac{8}{3}x-\frac{8}{3}+\frac{1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Since -\frac{8}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.

$$\frac{16}{3}-10x-\left(\frac{8}{3}x+\frac{-8+1}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

Add -8 and 1 to get -7.

$$\frac{16}{3}-10x-\left(\frac{8}{3}x-\frac{7}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

To find the opposite of \frac{8}{3}x-\frac{7}{3}, find the opposite of each term.

$$\frac{16}{3}-10x-\frac{8}{3}x-\left(-\frac{7}{3}\right)=-5x+\frac{2}{3}\left(x-1\right)$$

The opposite of -\frac{7}{3} is \frac{7}{3}.

$$\frac{16}{3}-10x-\frac{8}{3}x+\frac{7}{3}=-5x+\frac{2}{3}\left(x-1\right)$$

Combine -10x and -\frac{8}{3}x to get -\frac{38}{3}x.

$$\frac{16}{3}-\frac{38}{3}x+\frac{7}{3}=-5x+\frac{2}{3}\left(x-1\right)$$

Since \frac{16}{3} and \frac{7}{3} have the same denominator, add them by adding their numerators.

$$\frac{16+7}{3}-\frac{38}{3}x=-5x+\frac{2}{3}\left(x-1\right)$$

Add 16 and 7 to get 23.

$$\frac{23}{3}-\frac{38}{3}x=-5x+\frac{2}{3}\left(x-1\right)$$

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

$$\frac{23}{3}-\frac{38}{3}x=-5x+\frac{2}{3}x+\frac{2}{3}\left(-1\right)$$

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

$$\frac{23}{3}-\frac{38}{3}x=-5x+\frac{2}{3}x-\frac{2}{3}$$

Combine -5x and \frac{2}{3}x to get -\frac{13}{3}x.

$$\frac{23}{3}-\frac{38}{3}x=-\frac{13}{3}x-\frac{2}{3}$$

Add \frac{13}{3}x to both sides.

$$\frac{23}{3}-\frac{38}{3}x+\frac{13}{3}x=-\frac{2}{3}$$

Combine -\frac{38}{3}x and \frac{13}{3}x to get -\frac{25}{3}x.

$$\frac{23}{3}-\frac{25}{3}x=-\frac{2}{3}$$

Subtract \frac{23}{3} from both sides.

$$-\frac{25}{3}x=-\frac{2}{3}-\frac{23}{3}$$

Since -\frac{2}{3} and \frac{23}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{25}{3}x=\frac{-2-23}{3}$$

Subtract 23 from -2 to get -25.

$$-\frac{25}{3}x=-\frac{25}{3}$$

Multiply both sides by -\frac{3}{25}, the reciprocal of -\frac{25}{3}.

$$x=-\frac{25}{3}\left(-\frac{3}{25}\right)$$

Multiply -\frac{25}{3} times -\frac{3}{25} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{-25\left(-3\right)}{3\times 25}$$

Do the multiplications in the fraction \frac{-25\left(-3\right)}{3\times 25}.

$$x=\frac{75}{75}$$

Divide 75 by 75 to get 1.

$$x=1$$