Solve for x 1/2(6x+2)=5(x+3)

$\frac{1}{2} (6 x + 2) = 5 (x + 3)$

Step 1

Simplify $\frac{1}{2} \cdot (6 x + 2)$ .

Start simplifying.

$0 + 0 + \frac{1}{2} \cdot (6 x + 2) = 5 (x + 3)$

Simplify by adding zeros.

$\frac{1}{2} \cdot (6 x + 2) = 5 (x + 3)$

Apply the distributive property.

$\frac{1}{2} (6 x) + \frac{1}{2} \cdot 2 = 5 (x + 3)$

Cancel the common factor of $2$ .

Factor $2$ out of $6 x$ .

$\frac{1}{2} (2 (3 x)) + \frac{1}{2} \cdot 2 = 5 (x + 3)$

Cancel the common factor.

$\frac{1}{2} (2 (3 x)) + \frac{1}{2} \cdot 2 = 5 (x + 3)$

Rewrite the expression.

$3 x + \frac{1}{2} \cdot 2 = 5 (x + 3)$

Cancel the common factor of $2$ .

Cancel the common factor.

$3 x + \frac{1}{2} \cdot 2 = 5 (x + 3)$

Rewrite the expression.

$3 x + 1 = 5 (x + 3)$

Step 2

Simplify $5 (x + 3)$ .

Apply the distributive property.

$3 x + 1 = 5 x + 5 \cdot 3$

Multiply $5$ by $3$ .

$3 x + 1 = 5 x + 15$

Step 3

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

Subtract $5 x$ from both sides of the equation.

$3 x + 1 - 5 x = 15$

Subtract $5 x$ from $3 x$ .

$- 2 x + 1 = 15$

Step 4

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

Subtract $1$ from both sides of the equation.

$- 2 x = 15 - 1$

Subtract $1$ from $15$ .

$- 2 x = 14$

Step 5

Divide each term in $- 2 x = 14$ by $- 2$ and simplify.

Divide each term in $- 2 x = 14$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{14}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{14}{- 2}$

Divide $x$ by $1$ .

$x = \frac{14}{- 2}$

Simplify the right side.

Divide $14$ by $- 2$ .

$x = - 7$

Solve for x 1/2(6x+2)=5(x+3)

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