Solve 4x+1/frac{2{3}}-(1+4/3x)/frac{1{2}}=frac{5x-2/2}{1-frac{1}{2}}

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

Solve for x

$$x=\frac{9}{10}=0.9$$

Steps for Solving Linear Equation

Convert 1 to fraction \frac{2}{2}.

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

Since \frac{2}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 1 from 2 to get 1.

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

Divide each term of 4x+1 by \frac{2}{3} to get \frac{4x}{\frac{2}{3}}+\frac{1}{\frac{2}{3}}.

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

Divide 4x by \frac{2}{3} to get 6x.

$$6x+\frac{1}{\frac{2}{3}}-\frac{1+\frac{4}{3}x}{\frac{1}{2}}=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Divide 1 by \frac{2}{3} by multiplying 1 by the reciprocal of \frac{2}{3}.

$$6x+1\times \frac{3}{2}-\frac{1+\frac{4}{3}x}{\frac{1}{2}}=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

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

$$6x+\frac{3}{2}-\frac{1+\frac{4}{3}x}{\frac{1}{2}}=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Divide each term of 1+\frac{4}{3}x by \frac{1}{2} to get \frac{1}{\frac{1}{2}}+\frac{\frac{4}{3}x}{\frac{1}{2}}.

$$6x+\frac{3}{2}-\left(\frac{1}{\frac{1}{2}}+\frac{\frac{4}{3}x}{\frac{1}{2}}\right)=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Divide 1 by \frac{1}{2} by multiplying 1 by the reciprocal of \frac{1}{2}.

$$6x+\frac{3}{2}-\left(1\times 2+\frac{\frac{4}{3}x}{\frac{1}{2}}\right)=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Multiply 1 and 2 to get 2.

$$6x+\frac{3}{2}-\left(2+\frac{\frac{4}{3}x}{\frac{1}{2}}\right)=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Divide \frac{4}{3}x by \frac{1}{2} to get \frac{8}{3}x.

$$6x+\frac{3}{2}-\left(2+\frac{8}{3}x\right)=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

To find the opposite of 2+\frac{8}{3}x, find the opposite of each term.

$$6x+\frac{3}{2}-2-\frac{8}{3}x=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Convert 2 to fraction \frac{4}{2}.

$$6x+\frac{3}{2}-\frac{4}{2}-\frac{8}{3}x=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Since \frac{3}{2} and \frac{4}{2} have the same denominator, subtract them by subtracting their numerators.

$$6x+\frac{3-4}{2}-\frac{8}{3}x=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Subtract 4 from 3 to get -1.

$$6x-\frac{1}{2}-\frac{8}{3}x=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Combine 6x and -\frac{8}{3}x to get \frac{10}{3}x.

$$\frac{10}{3}x-\frac{1}{2}=\frac{\frac{5x-2}{2}}{\frac{1}{2}}$$

Divide each term of 5x-2 by 2 to get \frac{5}{2}x-1.

$$\frac{10}{3}x-\frac{1}{2}=\frac{\frac{5}{2}x-1}{\frac{1}{2}}$$

Divide each term of \frac{5}{2}x-1 by \frac{1}{2} to get -2+5x.

$$\frac{10}{3}x-\frac{1}{2}=-2+5x$$

Subtract 5x from both sides.

$$\frac{10}{3}x-\frac{1}{2}-5x=-2$$

Combine \frac{10}{3}x and -5x to get -\frac{5}{3}x.

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

Add \frac{1}{2} to both sides.

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

Convert -2 to fraction -\frac{4}{2}.

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

Since -\frac{4}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Add -4 and 1 to get -3.

$$-\frac{5}{3}x=-\frac{3}{2}$$

Multiply both sides by -\frac{3}{5}, the reciprocal of -\frac{5}{3}.

$$x=-\frac{3}{2}\left(-\frac{3}{5}\right)$$

Multiply -\frac{3}{2} times -\frac{3}{5} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{-3\left(-3\right)}{2\times 5}$$

Do the multiplications in the fraction \frac{-3\left(-3\right)}{2\times 5}.

$$x=\frac{9}{10}$$