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

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

Subtract $\frac{3 - x}{1 - x}$ from both sides of the inequality.

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

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

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

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

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

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

Write each expression with a common denominator of $(x + 1) (1 - x)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{x + 1}$ by $\frac{1 - x}{1 - x}$ .

$\frac{x (1 - x)}{(x + 1) (1 - x)} - \frac{3 - x}{1 - x} \cdot \frac{x + 1}{x + 1} < 0$

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

$\frac{x (1 - x)}{(x + 1) (1 - x)} - \frac{(3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Reorder the factors of $(x + 1) (1 - x)$ .

$\frac{x (1 - x)}{(1 - x) (x + 1)} - \frac{(3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Combine the numerators over the common denominator.

$\frac{x (1 - x) - (3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot 1 + x (- x) - (3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Multiply $x$ by $1$ .

$\frac{x + x (- x) - (3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Rewrite using the commutative property of multiplication.

$\frac{x - x \cdot x - (3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\frac{x - (x \cdot x) - (3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Multiply $x$ by $x$ .

$\frac{x - x^{2} - (3 - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Apply the distributive property.

$\frac{x - x^{2} + (- 1 \cdot 3 - - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Multiply $- 1$ by $3$ .

$\frac{x - x^{2} + (- 3 - - x) (x + 1)}{(1 - x) (x + 1)} < 0$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$\frac{x - x^{2} + (- 3 + 1 x) (x + 1)}{(1 - x) (x + 1)} < 0$

Multiply $x$ by $1$ .

$\frac{x - x^{2} + (- 3 + x) (x + 1)}{(1 - x) (x + 1)} < 0$

Expand $(- 3 + x) (x + 1)$ using the FOIL Method.

Apply the distributive property.

$\frac{x - x^{2} - 3 (x + 1) + x (x + 1)}{(1 - x) (x + 1)} < 0$

Apply the distributive property.

$\frac{x - x^{2} - 3 x - 3 \cdot 1 + x (x + 1)}{(1 - x) (x + 1)} < 0$

Apply the distributive property.

$\frac{x - x^{2} - 3 x - 3 \cdot 1 + x \cdot x + x \cdot 1}{(1 - x) (x + 1)} < 0$

Simplify and combine like terms.

Simplify each term.

Multiply $- 3$ by $1$ .

$\frac{x - x^{2} - 3 x - 3 + x \cdot x + x \cdot 1}{(1 - x) (x + 1)} < 0$

Multiply $x$ by $x$ .

$\frac{x - x^{2} - 3 x - 3 + x^{2} + x \cdot 1}{(1 - x) (x + 1)} < 0$

Multiply $x$ by $1$ .

$\frac{x - x^{2} - 3 x - 3 + x^{2} + x}{(1 - x) (x + 1)} < 0$

Add $- 3 x$ and $x$ .

$\frac{x - x^{2} - 2 x - 3 + x^{2}}{(1 - x) (x + 1)} < 0$

Subtract $2 x$ from $x$ .

$\frac{- x^{2} - x - 3 + x^{2}}{(1 - x) (x + 1)} < 0$

Add $- x^{2}$ and $x^{2}$ .

$\frac{- x - 3 + 0}{(1 - x) (x + 1)} < 0$

Add $- x - 3$ and $0$ .

$\frac{- x - 3}{(1 - x) (x + 1)} < 0$

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

$\frac{- (x) - 3}{(1 - x) (x + 1)} < 0$

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

$\frac{- (x) - 1 (3)}{(1 - x) (x + 1)} < 0$

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

$\frac{- (x + 3)}{(1 - x) (x + 1)} < 0$

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

$\frac{- 1 (x + 3)}{(1 - x) (x + 1)} < 0$

Move the negative in front of the fraction.

$- \frac{x + 3}{(1 - x) (x + 1)} < 0$

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

$x + 3 = 0$

$1 - x = 0$

$x + 1 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = - 3$

$x = 1$

$x = - 1$

Consolidate the solutions.

$x = - 3, 1, - 1$

Find the domain of $- \frac{x + 3}{(1 - x) (x + 1)}$ .

Set the denominator in $\frac{x + 3}{(1 - x) (x + 1)}$ equal to $0$ to find where the expression is undefined.

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

Solve for $x$ .

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$1 - x = 0$

$x + 1 = 0$

Set $1 - x$ equal to $0$ and solve for $x$ .

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(1 - x) (x + 1) = 0$ true.

$x = 1, - 1$

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

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$x > 1$

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 $x < - 3$ 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.

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

The left side $1.2$ is less than the right side $1.\overline{285714}$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

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

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

The left side $2$ is not less than the right side $1.\overline{6}$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{0}{(0) + 1} < \frac{3 - (0)}{1 - (0)}$

The left side $0$ is less than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\frac{4}{(4) + 1} < \frac{3 - (4)}{1 - (4)}$

The left side $0.8$ is not less than the right side $0.\overline{3}$ , which means that the given statement is false.

False

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

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$x < - 3$ or $- 1 < x < 1$

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 1 < x < 1$

Interval Notation:

$(- \infty, - 3) \cup (- 1, 1)$

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

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