Solve for n 1/12n+2=7/12n-3+5/6n

$\frac{1}{12} n + 2 = \frac{7}{12} n - 3 + \frac{5}{6} n$

Since $n$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\frac{7}{12} n - 3 + \frac{5}{6} n = \frac{1}{12} n + 2$

Simplify $\frac{7}{12} n - 3 + \frac{5}{6} n$ .

Simplify each term.

Combine $\frac{7}{12}$ and $n$ .

$\frac{7 n}{12} - 3 + \frac{5}{6} n = \frac{1}{12} n + 2$

Combine $\frac{5}{6}$ and $n$ .

$\frac{7 n}{12} - 3 + \frac{5 n}{6} = \frac{1}{12} n + 2$

To write $\frac{5 n}{6}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{7 n}{12} + \frac{5 n}{6} \cdot \frac{2}{2} - 3 = \frac{1}{12} n + 2$

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

Multiply $\frac{5 n}{6}$ by $\frac{2}{2}$ .

$\frac{7 n}{12} + \frac{5 n \cdot 2}{6 \cdot 2} - 3 = \frac{1}{12} n + 2$

Multiply $6$ by $2$ .

$\frac{7 n}{12} + \frac{5 n \cdot 2}{12} - 3 = \frac{1}{12} n + 2$

Combine the numerators over the common denominator.

$\frac{7 n + 5 n \cdot 2}{12} - 3 = \frac{1}{12} n + 2$

Simplify each term.

Simplify the numerator.

Factor $n$ out of $7 n + 5 n \cdot 2$ .

Factor $n$ out of $7 n$ .

$\frac{n \cdot 7 + 5 n \cdot 2}{12} - 3 = \frac{1}{12} n + 2$

Factor $n$ out of $5 n \cdot 2$ .

$\frac{n \cdot 7 + n (5 \cdot 2)}{12} - 3 = \frac{1}{12} n + 2$

Factor $n$ out of $n \cdot 7 + n (5 \cdot 2)$ .

$\frac{n (7 + 5 \cdot 2)}{12} - 3 = \frac{1}{12} n + 2$

Multiply $5$ by $2$ .

$\frac{n (7 + 10)}{12} - 3 = \frac{1}{12} n + 2$

Add $7$ and $10$ .

$\frac{n \cdot 17}{12} - 3 = \frac{1}{12} n + 2$

Move $17$ to the left of $n$ .

$\frac{17 n}{12} - 3 = \frac{1}{12} n + 2$

Combine $\frac{1}{12}$ and $n$ .

$\frac{17 n}{12} - 3 = \frac{n}{12} + 2$

Move all terms containing $n$ to the left side of the equation.

Subtract $\frac{n}{12}$ from both sides of the equation.

$\frac{17 n}{12} - 3 - \frac{n}{12} = 2$

Combine the numerators over the common denominator.

$- 3 + \frac{17 n - n}{12} = 2$

Subtract $n$ from $17 n$ .

$- 3 + \frac{16 n}{12} = 2$

Cancel the common factor of $16$ and $12$ .

Factor $4$ out of $16 n$ .

$- 3 + \frac{4 (4 n)}{12} = 2$

Cancel the common factors.

Factor $4$ out of $12$ .

$- 3 + \frac{4 (4 n)}{4 (3)} = 2$

Cancel the common factor.

$- 3 + \frac{4 (4 n)}{4 \cdot 3} = 2$

Rewrite the expression.

$- 3 + \frac{4 n}{3} = 2$

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

Add $3$ to both sides of the equation.

$\frac{4 n}{3} = 2 + 3$

Add $2$ and $3$ .

$\frac{4 n}{3} = 5$

Multiply both sides of the equation by $\frac{3}{4}$ .

$\frac{3}{4} \cdot \frac{4 n}{3} = \frac{3}{4} \cdot 5$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{3}{4} \cdot \frac{4 n}{3}$ .

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3}{4} \cdot \frac{4 n}{3} = \frac{3}{4} \cdot 5$

Rewrite the expression.

$\frac{1}{4} (4 n) = \frac{3}{4} \cdot 5$

Cancel the common factor of $4$ .

Factor $4$ out of $4 n$ .

$\frac{1}{4} (4 (n)) = \frac{3}{4} \cdot 5$

Cancel the common factor.

$\frac{1}{4} (4 n) = \frac{3}{4} \cdot 5$

Rewrite the expression.

$n = \frac{3}{4} \cdot 5$

Simplify the right side.

Multiply $\frac{3}{4} \cdot 5$ .

Combine $\frac{3}{4}$ and $5$ .

$n = \frac{3 \cdot 5}{4}$

Multiply $3$ by $5$ .

$n = \frac{15}{4}$

The result can be shown in multiple forms.

Exact Form:

$n = \frac{15}{4}$

Decimal Form:

$n = 3.75$

Mixed Number Form:

$n = 3 \frac{3}{4}$

Solve for n 1/12n+2=7/12n-3+5/6n

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