Solve for x |(2x-1)/(x+3)|=(1-2x)/(-x-3)

$| \frac{2 x - 1}{x + 3} | = \frac{1 - 2 x}{- x - 3}$

Multiply both sides by $- x - 3$ .

$| \frac{2 x - 1}{x + 3} | (- x - 3) = \frac{1 - 2 x}{- x - 3} (- x - 3)$

Simplify.

Simplify the left side.

Simplify $| \frac{2 x - 1}{x + 3} | (- x - 3)$ .

Apply the distributive property.

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

Simplify the expression.

Rewrite using the commutative property of multiplication.

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

Move $- 3$ to the left of $| \frac{2 x - 1}{x + 3} |$ .

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

Reorder factors in $- | \frac{2 x - 1}{x + 3} | x - 3 | \frac{2 x - 1}{x + 3} |$ .

$- x | \frac{2 x - 1}{x + 3} | - 3 | \frac{2 x - 1}{x + 3} | = \frac{1 - 2 x}{- x - 3} (- x - 3)$

Simplify the right side.

Simplify $\frac{1 - 2 x}{- x - 3} (- x - 3)$ .

Multiply $\frac{1 - 2 x}{- x - 3}$ by $- x - 3$ .

$- x | \frac{2 x - 1}{x + 3} | - 3 | \frac{2 x - 1}{x + 3} | = \frac{(1 - 2 x) (- x - 3)}{- x - 3}$

Cancel the common factor of $- x - 3$ .

Cancel the common factor.

$- x | \frac{2 x - 1}{x + 3} | - 3 | \frac{2 x - 1}{x + 3} | = \frac{(1 - 2 x) (- x - 3)}{- x - 3}$

Divide $1 - 2 x$ by $1$ .

$- x | \frac{2 x - 1}{x + 3} | - 3 | \frac{2 x - 1}{x + 3} | = 1 - 2 x$

Reorder $1$ and $- 2 x$ .

$- x | \frac{2 x - 1}{x + 3} | - 3 | \frac{2 x - 1}{x + 3} | = - 2 x + 1$

Solve for $x$ .

Factor $| \frac{2 x - 1}{x + 3} |$ out of $- x | \frac{2 x - 1}{x + 3} | - 3 | \frac{2 x - 1}{x + 3} |$ .

Factor $| \frac{2 x - 1}{x + 3} |$ out of $- x | \frac{2 x - 1}{x + 3} |$ .

$| \frac{2 x - 1}{x + 3} | (- x) - 3 | \frac{2 x - 1}{x + 3} | = - 2 x + 1$

Factor $| \frac{2 x - 1}{x + 3} |$ out of $- 3 | \frac{2 x - 1}{x + 3} |$ .

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

Factor $| \frac{2 x - 1}{x + 3} |$ out of $| \frac{2 x - 1}{x + 3} | (- x) + | \frac{2 x - 1}{x + 3} | \cdot - 3$ .

$| \frac{2 x - 1}{x + 3} | (- x - 3) = - 2 x + 1$

Divide each term in $| \frac{2 x - 1}{x + 3} | (- x - 3) = - 2 x + 1$ by $- x - 3$ and simplify.

Divide each term in $| \frac{2 x - 1}{x + 3} | (- x - 3) = - 2 x + 1$ by $- x - 3$ .

$\frac{| \frac{2 x - 1}{x + 3} | (- x - 3)}{- x - 3} = \frac{- 2 x}{- x - 3} + \frac{1}{- x - 3}$

Simplify the left side.

Cancel the common factor of $- x - 3$ .

Cancel the common factor.

$\frac{| \frac{2 x - 1}{x + 3} | (- x - 3)}{- x - 3} = \frac{- 2 x}{- x - 3} + \frac{1}{- x - 3}$

Divide $| \frac{2 x - 1}{x + 3} |$ by $1$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- 2 x}{- x - 3} + \frac{1}{- x - 3}$

Simplify the right side.

Combine the numerators over the common denominator.

$| \frac{2 x - 1}{x + 3} | = \frac{- 2 x + 1}{- x - 3}$

Factor $- 1$ out of $- 2 x$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- (2 x) + 1}{- x - 3}$

Rewrite $1$ as $- 1 (- 1)$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- (2 x) - 1 (- 1)}{- x - 3}$

Factor $- 1$ out of $- (2 x) - 1 (- 1)$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- (2 x - 1)}{- x - 3}$

Rewrite $- (2 x - 1)$ as $- 1 (2 x - 1)$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- 1 (2 x - 1)}{- x - 3}$

Factor $- 1$ out of $- x$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- 1 (2 x - 1)}{- (x) - 3}$

Rewrite $- 3$ as $- 1 (3)$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- 1 (2 x - 1)}{- (x) - 1 (3)}$

Factor $- 1$ out of $- (x) - 1 (3)$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- 1 (2 x - 1)}{- (x + 3)}$

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

$| \frac{2 x - 1}{x + 3} | = \frac{- 1 (2 x - 1)}{- 1 (x + 3)}$

Cancel the common factor.

$| \frac{2 x - 1}{x + 3} | = \frac{- 1 (2 x - 1)}{- 1 (x + 3)}$

Rewrite the expression.

$| \frac{2 x - 1}{x + 3} | = \frac{2 x - 1}{x + 3}$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$\frac{2 x - 1}{x + 3} = \pm \frac{2 x - 1}{x + 3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$\frac{2 x - 1}{x + 3} = \frac{2 x - 1}{x + 3}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$2 x - 1 = 2 x - 1$

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

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

$2 x - 1 - 2 x = - 1$

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

Subtract $2 x$ from $2 x$ .

$0 - 1 = - 1$

Subtract $1$ from $0$ .

$- 1 = - 1$

Since $- 1 = - 1$ , the equation will always be true.

All real numbers

Next, use the negative value of the $\pm$ to find the second solution.

$\frac{2 x - 1}{x + 3} = - \frac{2 x - 1}{x + 3}$

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

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

Simplify $- (2 x - 1)$ .

Rewrite.

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

Simplify by adding zeros.

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

Apply the distributive property.

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

Multiply.

Multiply $2$ by $- 1$ .

$2 x - 1 = - 2 x - - 1$

Multiply $- 1$ by $- 1$ .

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

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

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

$2 x - 1 + 2 x = 1$

Add $2 x$ and $2 x$ .

$4 x - 1 = 1$

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

Add $1$ to both sides of the equation.

$4 x = 1 + 1$

Add $1$ and $1$ .

$4 x = 2$

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 complete solution is the result of both the positive and negative portions of the solution.

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

Verify each of the solutions by substituting them into $| \frac{2 x - 1}{x + 3} | = \frac{1 - 2 x}{- x - 3}$ and solving.

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.5$

Solve for x |(2x-1)/(x+3)|=(1-2x)/(-x-3)

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