Solve for x |2x+1|<|2x-3|

$| 2 x + 1 | < | 2 x - 3 |$

Replace $<$ with $=$ in $| 2 x + 1 | < | 2 x - 3 |$ .

$| 2 x + 1 | = | 2 x - 3 |$

Rewrite the absolute value equation as four equations without absolute value bars.

$2 x + 1 = 2 x - 3$

$2 x + 1 = - (2 x - 3)$

$- (2 x + 1) = 2 x - 3$

$- (2 x + 1) = - (2 x - 3)$

After simplifying, there are only two unique equations to be solved.

$2 x + 1 = 2 x - 3$

$2 x + 1 = - (2 x - 3)$

Solve $2 x + 1 = 2 x - 3$ for $x$ .

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

Subtract $2 x$ from both sides of the equation.

$2 x + 1 - 2 x = - 3$

Combine the opposite terms in $2 x + 1 - 2 x$ .

Subtract $2 x$ from $2 x$ .

$0 + 1 = - 3$

Add $0$ and $1$ .

$1 = - 3$

Since $1 \neq - 3$ , there are no solutions.

No solution

Solve $2 x + 1 = - (2 x - 3)$ for $x$ .

Simplify $- (2 x - 3)$ .

Rewrite.

$2 x + 1 = 0 + 0 - (2 x - 3)$

Simplify by adding zeros.

$2 x + 1 = - (2 x - 3)$

Apply the distributive property.

$2 x + 1 = - (2 x) - - 3$

Multiply.

Multiply $2$ by $- 1$ .

$2 x + 1 = - 2 x - - 3$

Multiply $- 1$ by $- 3$ .

$2 x + 1 = - 2 x + 3$

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

Add $2 x$ to both sides of the equation.

$2 x + 1 + 2 x = 3$

Add $2 x$ and $2 x$ .

$4 x + 1 = 3$

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

Subtract $1$ from both sides of the equation.

$4 x = 3 - 1$

Subtract $1$ from $3$ .

$4 x = 2$

Divide each term in $4 x = 2$ by $4$ and simplify.

Divide each term in $4 x = 2$ by $4$ .

$\frac{4 x}{4} = \frac{2}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{2}{4}$

Divide $x$ by $1$ .

$x = \frac{2}{4}$

Simplify the right side.

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$x = \frac{2 (1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

$x = \frac{2 \cdot 1}{2 \cdot 2}$

Cancel the common factor.

$x = \frac{2 \cdot 1}{2 \cdot 2}$

Rewrite the expression.

$x = \frac{1}{2}$

The solution consists of all of the true intervals.

$x < \frac{1}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{1}{2}$

Interval Notation:

$(- \infty, \frac{1}{2})$

Solve for x |2x+1|<|2x-3|

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