Solve left(1/3+1/6right)y+4-y/6=1

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

Solve for y

$$y=1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 6, the least common multiple of 3,6.

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

Least common multiple of 3 and 6 is 6. Convert \frac{1}{3} and \frac{1}{6} to fractions with denominator 6.

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

Since \frac{2}{6} and \frac{1}{6} have the same denominator, add them by adding their numerators.

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

Add 2 and 1 to get 3.

$$6\times \frac{3}{6}y+4-y=6$$

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

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

Multiply 6 and \frac{1}{2} to get \frac{6}{2}.

$$\frac{6}{2}y+4-y=6$$

Divide 6 by 2 to get 3.

$$3y+4-y=6$$

Combine 3y and -y to get 2y.

$$2y+4=6$$

Subtract 4 from both sides.

$$2y=6-4$$

Subtract 4 from 6 to get 2.

$$2y=2$$

Divide both sides by 2.

$$y=\frac{2}{2}$$

Divide 2 by 2 to get 1.

$$y=1$$