Solve for c 5(2c-3)=3(3c-5)

$5 (2 c - 3) = 3 (3 c - 5)$

Step 1

Simplify $5 (2 c - 3)$ .

Start simplifying.

$0 + 0 + 5 (2 c - 3) = 3 (3 c - 5)$

Simplify by adding zeros.

$5 (2 c - 3) = 3 (3 c - 5)$

Apply the distributive property.

$5 (2 c) + 5 \cdot - 3 = 3 (3 c - 5)$

Multiply.

Multiply $2$ by $5$ .

$10 c + 5 \cdot - 3 = 3 (3 c - 5)$

Multiply $5$ by $- 3$ .

$10 c - 15 = 3 (3 c - 5)$

Step 2

Simplify $3 (3 c - 5)$ .

Apply the distributive property.

$10 c - 15 = 3 (3 c) + 3 \cdot - 5$

Multiply.

Multiply $3$ by $3$ .

$10 c - 15 = 9 c + 3 \cdot - 5$

Multiply $3$ by $- 5$ .

$10 c - 15 = 9 c - 15$

Step 3

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

Subtract $9 c$ from both sides of the equation.

$10 c - 15 - 9 c = - 15$

Subtract $9 c$ from $10 c$ .

$c - 15 = - 15$

Step 4

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

Add $15$ to both sides of the equation.

$c = - 15 + 15$

Add $- 15$ and $15$ .

$c = 0$