Solve the Inequality for x (3+x)/(3-x)>=1

$\frac{3 + x}{3 - x} \geq 1$

Subtract $1$ from both sides of the inequality.

$\frac{3 + x}{3 - x} - 1 \geq 0$

Simplify $\frac{3 + x}{3 - x} - 1$ .

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3 - x}{3 - x}$ .

$\frac{3 + x}{3 - x} - 1 \cdot \frac{3 - x}{3 - x} \geq 0$

Combine $- 1$ and $\frac{3 - x}{3 - x}$ .

$\frac{3 + x}{3 - x} + \frac{- (3 - x)}{3 - x} \geq 0$

Combine the numerators over the common denominator.

$\frac{3 + x - (3 - x)}{3 - x} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 + x - 1 \cdot 3 - - x}{3 - x} \geq 0$

Multiply $- 1$ by $3$ .

$\frac{3 + x - 3 - - x}{3 - x} \geq 0$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$\frac{3 + x - 3 + 1 x}{3 - x} \geq 0$

Multiply $x$ by $1$ .

$\frac{3 + x - 3 + x}{3 - x} \geq 0$

Subtract $3$ from $3$ .

$\frac{x + 0 + x}{3 - x} \geq 0$

Add $x$ and $0$ .

$\frac{x + x}{3 - x} \geq 0$

Add $x$ and $x$ .

$\frac{2 x}{3 - x} \geq 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$2 x = 0$

$3 - x = 0$

Divide each term in $2 x = 0$ by $2$ and simplify.

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x = 0$

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = 3$

Consolidate the solutions.

$x = 0, 3$

Find the domain of $\frac{2 x}{3 - x}$ .

Set the denominator in $\frac{2 x}{3 - x}$ equal to $0$ to find where the expression is undefined.

$3 - x = 0$

Solve for $x$ .

Subtract $3$ from both sides of the equation.

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$

The domain is all values of $x$ that make the expression defined.

$(- \infty, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < 0$

$0 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\frac{3 - 2}{3 - (- 2)} \geq 1$

The left side $0.2$ is less than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\frac{3 + 2}{3 - (2)} \geq 1$

The left side $5$ is greater than the right side $1$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{3 + 6}{3 - (6)} \geq 1$

The left side $- 3$ is less than the right side $1$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < 3$ True

$x > 3$ False

$x < 0$ False

$0 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$0 \leq x < 3$

The result can be shown in multiple forms.

Inequality Form:

$0 \leq x < 3$

Interval Notation:

$[0, 3)$

Solve the Inequality for x (3+x)/(3-x)>=1

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