Solve the Inequality for x (x+5)/(1-x)<=(3-x)/(x+3)

$\frac{x + 5}{1 - x} \leq \frac{3 - x}{x + 3}$

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

$\frac{x + 5}{1 - x} - \frac{3 - x}{x + 3} \leq 0$

Simplify $\frac{x + 5}{1 - x} - \frac{3 - x}{x + 3}$ .

To write $\frac{x + 5}{1 - x}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{x + 5}{1 - x} \cdot \frac{x + 3}{x + 3} - \frac{3 - x}{x + 3} \leq 0$

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

$\frac{x + 5}{1 - x} \cdot \frac{x + 3}{x + 3} - \frac{3 - x}{x + 3} \cdot \frac{1 - x}{1 - x} \leq 0$

Write each expression with a common denominator of $(1 - x) (x + 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x + 5}{1 - x}$ by $\frac{x + 3}{x + 3}$ .

$\frac{(x + 5) (x + 3)}{(1 - x) (x + 3)} - \frac{3 - x}{x + 3} \cdot \frac{1 - x}{1 - x} \leq 0$

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

$\frac{(x + 5) (x + 3)}{(1 - x) (x + 3)} - \frac{(3 - x) (1 - x)}{(x + 3) (1 - x)} \leq 0$

Reorder the factors of $(x + 3) (1 - x)$ .

$\frac{(x + 5) (x + 3)}{(1 - x) (x + 3)} - \frac{(3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Combine the numerators over the common denominator.

$\frac{(x + 5) (x + 3) - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Simplify the numerator.

Expand $(x + 5) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$\frac{x (x + 3) + 5 (x + 3) - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Apply the distributive property.

$\frac{x \cdot x + x \cdot 3 + 5 (x + 3) - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Apply the distributive property.

$\frac{x \cdot x + x \cdot 3 + 5 x + 5 \cdot 3 - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 3 + 5 x + 5 \cdot 3 - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Move $3$ to the left of $x$ .

$\frac{x^{2} + 3 \cdot x + 5 x + 5 \cdot 3 - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Multiply $5$ by $3$ .

$\frac{x^{2} + 3 x + 5 x + 15 - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Add $3 x$ and $5 x$ .

$\frac{x^{2} + 8 x + 15 - (3 - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 + (- 1 \cdot 3 - - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Multiply $- 1$ by $3$ .

$\frac{x^{2} + 8 x + 15 + (- 3 - - x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$\frac{x^{2} + 8 x + 15 + (- 3 + 1 x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

Multiply $x$ by $1$ .

$\frac{x^{2} + 8 x + 15 + (- 3 + x) (1 - x)}{(1 - x) (x + 3)} \leq 0$

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

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 - 3 (1 - x) + x (1 - x)}{(1 - x) (x + 3)} \leq 0$

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 - 3 \cdot 1 - 3 (- x) + x (1 - x)}{(1 - x) (x + 3)} \leq 0$

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 - 3 \cdot 1 - 3 (- x) + x \cdot 1 + x (- x)}{(1 - x) (x + 3)} \leq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $- 3$ by $1$ .

$\frac{x^{2} + 8 x + 15 - 3 - 3 (- x) + x \cdot 1 + x (- x)}{(1 - x) (x + 3)} \leq 0$

Multiply $- 1$ by $- 3$ .

$\frac{x^{2} + 8 x + 15 - 3 + 3 x + x \cdot 1 + x (- x)}{(1 - x) (x + 3)} \leq 0$

Multiply $x$ by $1$ .

$\frac{x^{2} + 8 x + 15 - 3 + 3 x + x + x (- x)}{(1 - x) (x + 3)} \leq 0$

Rewrite using the commutative property of multiplication.

$\frac{x^{2} + 8 x + 15 - 3 + 3 x + x - x \cdot x}{(1 - x) (x + 3)} \leq 0$

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

Move $x$ .

$\frac{x^{2} + 8 x + 15 - 3 + 3 x + x - (x \cdot x)}{(1 - x) (x + 3)} \leq 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + 8 x + 15 - 3 + 3 x + x - x^{2}}{(1 - x) (x + 3)} \leq 0$

Add $3 x$ and $x$ .

$\frac{x^{2} + 8 x + 15 - 3 + 4 x - x^{2}}{(1 - x) (x + 3)} \leq 0$

Subtract $x^{2}$ from $x^{2}$ .

$\frac{8 x + 15 - 3 + 4 x + 0}{(1 - x) (x + 3)} \leq 0$

Add $8 x + 15 - 3 + 4 x$ and $0$ .

$\frac{8 x + 15 - 3 + 4 x}{(1 - x) (x + 3)} \leq 0$

Add $8 x$ and $4 x$ .

$\frac{12 x + 15 - 3}{(1 - x) (x + 3)} \leq 0$

Subtract $3$ from $15$ .

$\frac{12 x + 12}{(1 - x) (x + 3)} \leq 0$

Factor $12$ out of $12 x + 12$ .

Factor $12$ out of $12 x$ .

$\frac{12 (x) + 12}{(1 - x) (x + 3)} \leq 0$

Factor $12$ out of $12$ .

$\frac{12 (x) + 12 (1)}{(1 - x) (x + 3)} \leq 0$

Factor $12$ out of $12 (x) + 12 (1)$ .

$\frac{12 (x + 1)}{(1 - x) (x + 3)} \leq 0$

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

$x + 1 = 0$

$1 - x = 0$

$x + 3 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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 $3$ from both sides of the equation.

$x = - 3$

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

$x = - 1$

$x = 1$

$x = - 3$

Consolidate the solutions.

$x = - 1, 1, - 3$

Find the domain of $\frac{12 (x + 1)}{(1 - x) (x + 3)}$ .

Set the denominator in $\frac{12 (x + 1)}{(1 - x) (x + 3)}$ equal to $0$ to find where the expression is undefined.

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

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 $0$ .

$1 - x = 0$

$x + 3 = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(1 - x) (x + 3) = 0$ true.

$x = 1, - 3$

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

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

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.

$\frac{(- 6) + 5}{1 - (- 6)} \leq \frac{3 - (- 6)}{(- 6) + 3}$

The left side $- 0.\overline{142857}$ is greater than the right side $- 3$ , which means that the given statement is false.

False

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.

$\frac{(- 2) + 5}{1 - (- 2)} \leq \frac{3 - (- 2)}{(- 2) + 3}$

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

True

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.

$\frac{(0) + 5}{1 - (0)} \leq \frac{3 - (0)}{(0) + 3}$

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

False

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.

$\frac{(4) + 5}{1 - (4)} \leq \frac{3 - (4)}{(4) + 3}$

The left side $- 3$ is less than the right side $- 0.\overline{142857}$ , which means that the given statement is always true.

True

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

$x < - 3$ False

$- 3 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 3$ False

$- 3 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$- 3 < x \leq - 1$ or $x > 1$

The result can be shown in multiple forms.

Inequality Form:

$- 3 < x \leq - 1 or x > 1$

Interval Notation:

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

Solve the Inequality for x (x+5)/(1-x)<=(3-x)/(x+3)

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