Solve for y (4x+6y)/3=y+15

$\frac{4 x + 6 y}{3} = y + 15$

Step 1

Multiply both sides by $3$ .

$\frac{4 x + 6 y}{3} \cdot 3 = (y + 15) \cdot 3$

Step 2

Simplify.

Simplify the left side.

Simplify $\frac{4 x + 6 y}{3} \cdot 3$ .

Factor $2$ out of $4 x + 6 y$ .

Factor $2$ out of $4 x$ .

$\frac{2 (2 x) + 6 y}{3} \cdot 3 = (y + 15) \cdot 3$

Factor $2$ out of $6 y$ .

$\frac{2 (2 x) + 2 (3 y)}{3} \cdot 3 = (y + 15) \cdot 3$

Factor $2$ out of $2 (2 x) + 2 (3 y)$ .

$\frac{2 (2 x + 3 y)}{3} \cdot 3 = (y + 15) \cdot 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{2 (2 x + 3 y)}{3} \cdot 3 = (y + 15) \cdot 3$

Rewrite the expression.

$2 (2 x + 3 y) = (y + 15) \cdot 3$

Apply the distributive property.

$2 (2 x) + 2 (3 y) = (y + 15) \cdot 3$

Multiply.

Multiply $2$ by $2$ .

$4 x + 2 (3 y) = (y + 15) \cdot 3$

Multiply $3$ by $2$ .

$4 x + 6 y = (y + 15) \cdot 3$

Simplify the right side.

Simplify $(y + 15) \cdot 3$ .

Apply the distributive property.

$4 x + 6 y = y \cdot 3 + 15 \cdot 3$

Simplify the expression.

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

$4 x + 6 y = 3 \cdot y + 15 \cdot 3$

Multiply $15$ by $3$ .

$4 x + 6 y = 3 y + 45$

Step 3

Solve for $y$ .

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

Subtract $3 y$ from both sides of the equation.

$4 x + 6 y - 3 y = 45$

Subtract $3 y$ from $6 y$ .

$4 x + 3 y = 45$

Subtract $4 x$ from both sides of the equation.

$3 y = 45 - 4 x$

Divide each term in $3 y = 45 - 4 x$ by $3$ and simplify.

Divide each term in $3 y = 45 - 4 x$ by $3$ .

$\frac{3 y}{3} = \frac{45}{3} + \frac{- 4 x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{45}{3} + \frac{- 4 x}{3}$

Divide $y$ by $1$ .

$y = \frac{45}{3} + \frac{- 4 x}{3}$

Simplify the right side.

Simplify each term.

Divide $45$ by $3$ .

$y = 15 + \frac{- 4 x}{3}$

Move the negative in front of the fraction.

$y = 15 - \frac{4 x}{3}$

Solve for y (4x+6y)/3=y+15

Related Questions

Support LCMGCF.com by disabling your adblocker.