Solve the Inequality for x 2x<=(4x+1)/2

$2 x \leq \frac{4 x + 1}{2}$

Step 1

Multiply both sides by $2$ .

$2 x \cdot 2 \leq \frac{4 x + 1}{2} \cdot 2$

Step 2

Simplify.

Simplify the left side.

Multiply $2$ by $2$ .

$4 x \leq \frac{4 x + 1}{2} \cdot 2$

Simplify the right side.

Cancel the common factor of $2$ .

Cancel the common factor.

$4 x \leq \frac{4 x + 1}{2} \cdot 2$

Rewrite the expression.

$4 x \leq 4 x + 1$

Step 3

Solve for $x$ .

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

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

$4 x - 4 x \leq 1$

Subtract $4 x$ from $4 x$ .

$0 \leq 1$

Since $0 \leq 1$ , the inequality will always be true.

Always true

Step 4

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$