Solve 1-6-x/3/2+x-x/2-frac{3+x/4}{4}=3

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

Solve for x

$$x=3$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$2\left(1-\frac{6-x}{3}\right)+4x-\left(\frac{x}{2}-\frac{3+x}{4}\right)=12$$

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

$$2+2\left(-\frac{6-x}{3}\right)+4x-\left(\frac{x}{2}-\frac{3+x}{4}\right)=12$$

Express 2\left(-\frac{6-x}{3}\right) as a single fraction.

$$2+\frac{-2\left(6-x\right)}{3}+4x-\left(\frac{x}{2}-\frac{3+x}{4}\right)=12$$

Use the distributive property to multiply -2 by 6-x.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 4 is 4. Multiply \frac{x}{2} times \frac{2}{2}.

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

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

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

Do the multiplications in 2x-\left(3+x\right).

$$2+\frac{-12+2x}{3}+4x-\frac{2x-3-x}{4}=12$$

Combine like terms in 2x-3-x.

$$2+\frac{-12+2x}{3}+4x-\frac{x-3}{4}=12$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 4 is 12. Multiply \frac{-12+2x}{3} times \frac{4}{4}. Multiply \frac{x-3}{4} times \frac{3}{3}.

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

Since \frac{4\left(-12+2x\right)}{12} and \frac{3\left(x-3\right)}{12} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 4\left(-12+2x\right)-3\left(x-3\right).

$$2+\frac{-48+8x-3x+9}{12}+4x=12$$

Combine like terms in -48+8x-3x+9.

$$2+\frac{-39+5x}{12}+4x=12$$

Divide each term of -39+5x by 12 to get -\frac{13}{4}+\frac{5}{12}x.

$$2-\frac{13}{4}+\frac{5}{12}x+4x=12$$

Convert 2 to fraction \frac{8}{4}.

$$\frac{8}{4}-\frac{13}{4}+\frac{5}{12}x+4x=12$$

Since \frac{8}{4} and \frac{13}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{8-13}{4}+\frac{5}{12}x+4x=12$$

Subtract 13 from 8 to get -5.

$$-\frac{5}{4}+\frac{5}{12}x+4x=12$$

Combine \frac{5}{12}x and 4x to get \frac{53}{12}x.

$$-\frac{5}{4}+\frac{53}{12}x=12$$

Add \frac{5}{4} to both sides.

$$\frac{53}{12}x=12+\frac{5}{4}$$

Convert 12 to fraction \frac{48}{4}.

$$\frac{53}{12}x=\frac{48}{4}+\frac{5}{4}$$

Since \frac{48}{4} and \frac{5}{4} have the same denominator, add them by adding their numerators.

$$\frac{53}{12}x=\frac{48+5}{4}$$

Add 48 and 5 to get 53.

$$\frac{53}{12}x=\frac{53}{4}$$

Multiply both sides by \frac{12}{53}, the reciprocal of \frac{53}{12}.

$$x=\frac{53}{4}\times \frac{12}{53}$$

Multiply \frac{53}{4} times \frac{12}{53} by multiplying numerator times numerator and denominator times denominator.

$$x=\frac{53\times 12}{4\times 53}$$

Cancel out 53 in both numerator and denominator.

$$x=\frac{12}{4}$$

Divide 12 by 4 to get 3.

$$x=3$$