Solve 3/4(4/3(1/4x-1)+8]=7/3+frac{2}{3}x

$$\frac { 3 } { 4 } ( \frac { 4 } { 3 } ( \frac { 1 } { 4 } x - 1 ) + 8 ] = \frac { 7 } { 3 } + \frac { 2 } { 3 } x$$

Solve for x

$$x = \frac{32}{5} = 6\frac{2}{5} = 6.4$$

Steps for Solving Linear Equation

Use the distributive property to multiply \frac{4}{3} by \frac{1}{4}x-1.

$$\frac{3}{4}\left(\frac{4}{3}\times \frac{1}{4}x+\frac{4}{3}\left(-1\right)+8\right)=\frac{7}{3}+\frac{2}{3}x$$

Multiply \frac{4}{3} times \frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3}{4}\left(\frac{4\times 1}{3\times 4}x+\frac{4}{3}\left(-1\right)+8\right)=\frac{7}{3}+\frac{2}{3}x$$

Cancel out 4 in both numerator and denominator.

$$\frac{3}{4}\left(\frac{1}{3}x+\frac{4}{3}\left(-1\right)+8\right)=\frac{7}{3}+\frac{2}{3}x$$

Multiply \frac{4}{3} and -1 to get -\frac{4}{3}.

$$\frac{3}{4}\left(\frac{1}{3}x-\frac{4}{3}+8\right)=\frac{7}{3}+\frac{2}{3}x$$

Convert 8 to fraction \frac{24}{3}.

$$\frac{3}{4}\left(\frac{1}{3}x-\frac{4}{3}+\frac{24}{3}\right)=\frac{7}{3}+\frac{2}{3}x$$

Since -\frac{4}{3} and \frac{24}{3} have the same denominator, add them by adding their numerators.

$$\frac{3}{4}\left(\frac{1}{3}x+\frac{-4+24}{3}\right)=\frac{7}{3}+\frac{2}{3}x$$

Add -4 and 24 to get 20.

$$\frac{3}{4}\left(\frac{1}{3}x+\frac{20}{3}\right)=\frac{7}{3}+\frac{2}{3}x$$

Use the distributive property to multiply \frac{3}{4} by \frac{1}{3}x+\frac{20}{3}.

$$\frac{3}{4}\times \frac{1}{3}x+\frac{3}{4}\times \frac{20}{3}=\frac{7}{3}+\frac{2}{3}x$$

Multiply \frac{3}{4} times \frac{1}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{3\times 1}{4\times 3}x+\frac{3}{4}\times \frac{20}{3}=\frac{7}{3}+\frac{2}{3}x$$

Cancel out 3 in both numerator and denominator.

$$\frac{1}{4}x+\frac{3}{4}\times \frac{20}{3}=\frac{7}{3}+\frac{2}{3}x$$

Multiply \frac{3}{4} times \frac{20}{3} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{4}x+\frac{3\times 20}{4\times 3}=\frac{7}{3}+\frac{2}{3}x$$

Cancel out 3 in both numerator and denominator.

$$\frac{1}{4}x+\frac{20}{4}=\frac{7}{3}+\frac{2}{3}x$$

Divide 20 by 4 to get 5.

$$\frac{1}{4}x+5=\frac{7}{3}+\frac{2}{3}x$$

Subtract \frac{2}{3}x from both sides.

$$\frac{1}{4}x+5-\frac{2}{3}x=\frac{7}{3}$$

Combine \frac{1}{4}x and -\frac{2}{3}x to get -\frac{5}{12}x.

$$-\frac{5}{12}x+5=\frac{7}{3}$$

Subtract 5 from both sides.

$$-\frac{5}{12}x=\frac{7}{3}-5$$

Convert 5 to fraction \frac{15}{3}.

$$-\frac{5}{12}x=\frac{7}{3}-\frac{15}{3}$$

Since \frac{7}{3} and \frac{15}{3} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{5}{12}x=\frac{7-15}{3}$$

Subtract 15 from 7 to get -8.

$$-\frac{5}{12}x=-\frac{8}{3}$$

Multiply both sides by -\frac{12}{5}, the reciprocal of -\frac{5}{12}.

$$x=-\frac{8}{3}\left(-\frac{12}{5}\right)$$

Multiply -\frac{8}{3} times -\frac{12}{5} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{-8\left(-12\right)}{3\times 5}$$

Do the multiplications in the fraction \frac{-8\left(-12\right)}{3\times 5}.

$$x=\frac{96}{15}$$

Reduce the fraction \frac{96}{15} to lowest terms by extracting and canceling out 3.

$$x=\frac{32}{5}$$