Solve for c 2/3c-3+5/4c=-c+5/2c+2

$\frac{2}{3} c - 3 + \frac{5}{4} c = - c + \frac{5}{2} c + 2$

Step 1

Simplify $\frac{2}{3} c - 3 + \frac{5}{4} c$ .

Simplify each term.

Combine $\frac{2}{3}$ and $c$ .

$\frac{2 c}{3} - 3 + \frac{5}{4} c = - c + \frac{5}{2} c + 2$

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

$\frac{2 c}{3} - 3 + \frac{5 c}{4} = - c + \frac{5}{2} c + 2$

To write $\frac{2 c}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{2 c}{3} \cdot \frac{4}{4} + \frac{5 c}{4} - 3 = - c + \frac{5}{2} c + 2$

To write $\frac{5 c}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{2 c}{3} \cdot \frac{4}{4} + \frac{5 c}{4} \cdot \frac{3}{3} - 3 = - c + \frac{5}{2} c + 2$

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

Multiply $\frac{2 c}{3}$ by $\frac{4}{4}$ .

$\frac{2 c \cdot 4}{3 \cdot 4} + \frac{5 c}{4} \cdot \frac{3}{3} - 3 = - c + \frac{5}{2} c + 2$

Multiply $3$ by $4$ .

$\frac{2 c \cdot 4}{12} + \frac{5 c}{4} \cdot \frac{3}{3} - 3 = - c + \frac{5}{2} c + 2$

Multiply $\frac{5 c}{4}$ by $\frac{3}{3}$ .

$\frac{2 c \cdot 4}{12} + \frac{5 c \cdot 3}{4 \cdot 3} - 3 = - c + \frac{5}{2} c + 2$

Multiply $4$ by $3$ .

$\frac{2 c \cdot 4}{12} + \frac{5 c \cdot 3}{12} - 3 = - c + \frac{5}{2} c + 2$

Combine the numerators over the common denominator.

$\frac{2 c \cdot 4 + 5 c \cdot 3}{12} - 3 = - c + \frac{5}{2} c + 2$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $2 c \cdot 4 + 5 c \cdot 3$ .

Factor $c$ out of $2 c \cdot 4$ .

$\frac{c (2 \cdot 4) + 5 c \cdot 3}{12} - 3 = - c + \frac{5}{2} c + 2$

Factor $c$ out of $5 c \cdot 3$ .

$\frac{c (2 \cdot 4) + c (5 \cdot 3)}{12} - 3 = - c + \frac{5}{2} c + 2$

Factor $c$ out of $c (2 \cdot 4) + c (5 \cdot 3)$ .

$\frac{c (2 \cdot 4 + 5 \cdot 3)}{12} - 3 = - c + \frac{5}{2} c + 2$

Multiply $2$ by $4$ .

$\frac{c (8 + 5 \cdot 3)}{12} - 3 = - c + \frac{5}{2} c + 2$

Multiply $5$ by $3$ .

$\frac{c (8 + 15)}{12} - 3 = - c + \frac{5}{2} c + 2$

Add $8$ and $15$ .

$\frac{c \cdot 23}{12} - 3 = - c + \frac{5}{2} c + 2$

Move $23$ to the left of $c$ .

$\frac{23 c}{12} - 3 = - c + \frac{5}{2} c + 2$

Step 2

Simplify $- c + \frac{5}{2} c + 2$ .

Combine $\frac{5}{2}$ and $c$ .

$\frac{23 c}{12} - 3 = - c + \frac{5 c}{2} + 2$

To write $- c$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{23 c}{12} - 3 = - c \cdot \frac{2}{2} + \frac{5 c}{2} + 2$

Simplify terms.

Combine $- c$ and $\frac{2}{2}$ .

$\frac{23 c}{12} - 3 = \frac{- c \cdot 2}{2} + \frac{5 c}{2} + 2$

Combine the numerators over the common denominator.

$\frac{23 c}{12} - 3 = \frac{- c \cdot 2 + 5 c}{2} + 2$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $- c \cdot 2 + 5 c$ .

Factor $c$ out of $- c \cdot 2$ .

$\frac{23 c}{12} - 3 = \frac{c (- 1 \cdot 2) + 5 c}{2} + 2$

Factor $c$ out of $5 c$ .

$\frac{23 c}{12} - 3 = \frac{c (- 1 \cdot 2) + c \cdot 5}{2} + 2$

Factor $c$ out of $c (- 1 \cdot 2) + c \cdot 5$ .

$\frac{23 c}{12} - 3 = \frac{c (- 1 \cdot 2 + 5)}{2} + 2$

Multiply $- 1$ by $2$ .

$\frac{23 c}{12} - 3 = \frac{c (- 2 + 5)}{2} + 2$

Add $- 2$ and $5$ .

$\frac{23 c}{12} - 3 = \frac{c \cdot 3}{2} + 2$

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

$\frac{23 c}{12} - 3 = \frac{3 c}{2} + 2$

Step 3

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

Subtract $\frac{3 c}{2}$ from both sides of the equation.

$\frac{23 c}{12} - 3 - \frac{3 c}{2} = 2$

To write $- \frac{3 c}{2}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{23 c}{12} - \frac{3 c}{2} \cdot \frac{6}{6} - 3 = 2$

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

Multiply $\frac{3 c}{2}$ by $\frac{6}{6}$ .

$\frac{23 c}{12} - \frac{3 c \cdot 6}{2 \cdot 6} - 3 = 2$

Multiply $2$ by $6$ .

$\frac{23 c}{12} - \frac{3 c \cdot 6}{12} - 3 = 2$

Combine the numerators over the common denominator.

$\frac{23 c - 3 c \cdot 6}{12} - 3 = 2$

Simplify each term.

Simplify the numerator.

Factor $c$ out of $23 c - 3 c \cdot 6$ .

Factor $c$ out of $23 c$ .

$\frac{c \cdot 23 - 3 c \cdot 6}{12} - 3 = 2$

Factor $c$ out of $- 3 c \cdot 6$ .

$\frac{c \cdot 23 + c (- 3 \cdot 6)}{12} - 3 = 2$

Factor $c$ out of $c \cdot 23 + c (- 3 \cdot 6)$ .

$\frac{c (23 - 3 \cdot 6)}{12} - 3 = 2$

Multiply $- 3$ by $6$ .

$\frac{c (23 - 18)}{12} - 3 = 2$

Subtract $18$ from $23$ .

$\frac{c \cdot 5}{12} - 3 = 2$

Move $5$ to the left of $c$ .

$\frac{5 c}{12} - 3 = 2$

Step 4

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

Add $3$ to both sides of the equation.

$\frac{5 c}{12} = 2 + 3$

Add $2$ and $3$ .

$\frac{5 c}{12} = 5$

Step 5

Multiply both sides of the equation by $\frac{12}{5}$ .

$\frac{12}{5} \cdot \frac{5 c}{12} = \frac{12}{5} \cdot 5$

Step 6

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{12}{5} \cdot \frac{5 c}{12}$ .

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12}{5} \cdot \frac{5 c}{12} = \frac{12}{5} \cdot 5$

Rewrite the expression.

$\frac{1}{5} (5 c) = \frac{12}{5} \cdot 5$

Cancel the common factor of $5$ .

Factor $5$ out of $5 c$ .

$\frac{1}{5} (5 (c)) = \frac{12}{5} \cdot 5$

Cancel the common factor.

$\frac{1}{5} (5 c) = \frac{12}{5} \cdot 5$

Rewrite the expression.

$c = \frac{12}{5} \cdot 5$

Simplify the right side.

Cancel the common factor of $5$ .

Cancel the common factor.

$c = \frac{12}{5} \cdot 5$

Rewrite the expression.

$c = 12$

Solve for c 2/3c-3+5/4c=-c+5/2c+2

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