Solve for y (y-1)/4+1/3=7/12

$\frac{y - 1}{4} + \frac{1}{3} = \frac{7}{12}$

Simplify $\frac{y - 1}{4} + \frac{1}{3}$ .

To write $\frac{y - 1}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{y - 1}{4} \cdot \frac{3}{3} + \frac{1}{3} = \frac{7}{12}$

To write $\frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{y - 1}{4} \cdot \frac{3}{3} + \frac{1}{3} \cdot \frac{4}{4} = \frac{7}{12}$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{y - 1}{4}$ by $\frac{3}{3}$ .

$\frac{(y - 1) \cdot 3}{4 \cdot 3} + \frac{1}{3} \cdot \frac{4}{4} = \frac{7}{12}$

Multiply $4$ by $3$ .

$\frac{(y - 1) \cdot 3}{12} + \frac{1}{3} \cdot \frac{4}{4} = \frac{7}{12}$

Multiply $\frac{1}{3}$ by $\frac{4}{4}$ .

$\frac{(y - 1) \cdot 3}{12} + \frac{4}{3 \cdot 4} = \frac{7}{12}$

Multiply $3$ by $4$ .

$\frac{(y - 1) \cdot 3}{12} + \frac{4}{12} = \frac{7}{12}$

Combine the numerators over the common denominator.

$\frac{(y - 1) \cdot 3 + 4}{12} = \frac{7}{12}$

Simplify the numerator.

Apply the distributive property.

$\frac{y \cdot 3 - 1 \cdot 3 + 4}{12} = \frac{7}{12}$

Move $3$ to the left of $y$ .

$\frac{3 \cdot y - 1 \cdot 3 + 4}{12} = \frac{7}{12}$

Multiply $- 1$ by $3$ .

$\frac{3 \cdot y - 3 + 4}{12} = \frac{7}{12}$

Add $- 3$ and $4$ .

$\frac{3 y + 1}{12} = \frac{7}{12}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$3 y + 1 = 7$

Move all terms not containing $y$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$3 y = 7 - 1$

Subtract $1$ from $7$ .

$3 y = 6$

Divide each term in $3 y = 6$ by $3$ and simplify.

Divide each term in $3 y = 6$ by $3$ .

$\frac{3 y}{3} = \frac{6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{6}{3}$

Divide $y$ by $1$ .

$y = \frac{6}{3}$

Simplify the right side.

Divide $6$ by $3$ .

$y = 2$