Solve -(1-1/4x)div3/2+(x/3-2)div(-2/3)=2-x+frac{2x+1}{3}

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

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

Multiply both sides of the equation by 3.

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

To find the opposite of 1-\frac{1}{4}x, find the opposite of each term.

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

The opposite of -\frac{1}{4}x is \frac{1}{4}x.

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

Combine -3x and 2x to get -x.

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

Add 6 and 1 to get 7.

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

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

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

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

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

Divide \frac{1}{4}x by \frac{3}{2} to get \frac{1}{6}x.

$$3\left(-\frac{2}{3}+\frac{1}{6}x\right)+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Use the distributive property to multiply 3 by -\frac{2}{3}+\frac{1}{6}x.

$$3\left(-\frac{2}{3}\right)+3\times \frac{1}{6}x+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Cancel out 3 and 3.

$$-2+3\times \frac{1}{6}x+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Multiply 3 and \frac{1}{6} to get \frac{3}{6}.

$$-2+\frac{3}{6}x+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

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

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

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

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

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

Multiply -2 times -\frac{3}{2}.

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

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

$$-2+\frac{1}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}+9=7-x$$

Add -2 and 9 to get 7.

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

Add x to both sides.

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

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

$$7+\frac{3}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}=7$$

Subtract 7 from both sides.

$$\frac{3}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}=7-7$$

Subtract 7 from 7 to get 0.

$$\frac{3}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}=0$$

Reorder the terms.

$$3\times \frac{x}{-\frac{2}{3}\times 3}+\frac{3}{2}x=0$$

Cancel out 3 and 3.

$$3\times \frac{x}{-2}+\frac{3}{2}x=0$$

Express 3\times \frac{x}{-2} as a single fraction.

$$\frac{3x}{-2}+\frac{3}{2}x=0$$

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

$$-3x+3x=0$$

Combine -3x and 3x to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Multiply both sides of the equation by 3.

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

To find the opposite of 1-\frac{1}{4}x, find the opposite of each term.

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

The opposite of -\frac{1}{4}x is \frac{1}{4}x.

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

Combine -3x and 2x to get -x.

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

Add 6 and 1 to get 7.

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

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

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

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

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

Divide \frac{1}{4}x by \frac{3}{2} to get \frac{1}{6}x.

$$3\left(-\frac{2}{3}+\frac{1}{6}x\right)+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Use the distributive property to multiply 3 by -\frac{2}{3}+\frac{1}{6}x.

$$3\left(-\frac{2}{3}\right)+3\times \frac{1}{6}x+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Cancel out 3 and 3.

$$-2+3\times \frac{1}{6}x+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Multiply 3 and \frac{1}{6} to get \frac{3}{6}.

$$-2+\frac{3}{6}x+3\times \frac{\frac{x}{3}-2}{-\frac{2}{3}}=7-x$$

Reduce the fraction \frac{3}{6} to lowest terms by extracting and canceling out 3.

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

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

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

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

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

Multiply -2 times -\frac{3}{2}.

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

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

$$-2+\frac{1}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}+9=7-x$$

Add -2 and 9 to get 7.

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

Add x to both sides.

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

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

$$7+\frac{3}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}=7$$

Subtract 7 from both sides.

$$\frac{3}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}=7-7$$

Subtract 7 from 7 to get 0.

$$\frac{3}{2}x+3\times \frac{\frac{x}{3}}{-\frac{2}{3}}=0$$

Reorder the terms.

$$3\times \frac{x}{-\frac{2}{3}\times 3}+\frac{3}{2}x=0$$

Cancel out 3 and 3.

$$3\times \frac{x}{-2}+\frac{3}{2}x=0$$

Express 3\times \frac{x}{-2} as a single fraction.

$$\frac{3x}{-2}+\frac{3}{2}x=0$$

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

$$-3x+3x=0$$

Combine -3x and 3x to get 0.

$$0=0$$

Compare 0 and 0.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$