Solve for x 1/2(6x+20)=x+4+2(x+3)

$\frac{1}{2} (6 x + 20) = x + 4 + 2 (x + 3)$

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

$x + 4 + 2 (x + 3) = \frac{1}{2} \cdot (6 x + 20)$

Simplify $x + 4 + 2 (x + 3)$ .

Simplify each term.

Apply the distributive property.

$x + 4 + 2 x + 2 \cdot 3 = \frac{1}{2} \cdot (6 x + 20)$

Multiply $2$ by $3$ .

$x + 4 + 2 x + 6 = \frac{1}{2} \cdot (6 x + 20)$

Simplify by adding terms.

Add $x$ and $2 x$ .

$3 x + 4 + 6 = \frac{1}{2} \cdot (6 x + 20)$

Add $4$ and $6$ .

$3 x + 10 = \frac{1}{2} \cdot (6 x + 20)$

Simplify $\frac{1}{2} \cdot (6 x + 20)$ .

Apply the distributive property.

$3 x + 10 = \frac{1}{2} (6 x) + \frac{1}{2} \cdot 20$

Cancel the common factor of $2$ .

Factor $2$ out of $6 x$ .

$3 x + 10 = \frac{1}{2} (2 (3 x)) + \frac{1}{2} \cdot 20$

Cancel the common factor.

$3 x + 10 = \frac{1}{2} (2 (3 x)) + \frac{1}{2} \cdot 20$

Rewrite the expression.

$3 x + 10 = 3 x + \frac{1}{2} \cdot 20$

Cancel the common factor of $2$ .

Factor $2$ out of $20$ .

$3 x + 10 = 3 x + \frac{1}{2} \cdot (2 (10))$

Cancel the common factor.

$3 x + 10 = 3 x + \frac{1}{2} \cdot (2 \cdot 10)$

Rewrite the expression.

$3 x + 10 = 3 x + 10$

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

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

$3 x + 10 - 3 x = 10$

Combine the opposite terms in $3 x + 10 - 3 x$ .

Subtract $3 x$ from $3 x$ .

$0 + 10 = 10$

Add $0$ and $10$ .

$10 = 10$

Since $10 = 10$ , the equation will always be true for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$