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

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

Step 1

Subtract $\frac{x - 3}{x + 5}$ from both sides of the inequality.

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

Step 2

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

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

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

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

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

Write each expression with a common denominator of $(x - 1) (x + 5)$ , by multiplying each by an appropriate factor of $1$ .

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

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

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

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

Reorder the factors of $(x - 1) (x + 5)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $5$ to the left of $x$ .

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

Multiply $3$ by $5$ .

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

Add $5 x$ and $3 x$ .

$\frac{x^{2} + 8 x + 15 - (x - 3) (x - 1)}{(x + 5) (x - 1)} \leq 0$

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 + (- x - - 3) (x - 1)}{(x + 5) (x - 1)} \leq 0$

Multiply $- 1$ by $- 3$ .

$\frac{x^{2} + 8 x + 15 + (- x + 3) (x - 1)}{(x + 5) (x - 1)} \leq 0$

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

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 - x (x - 1) + 3 (x - 1)}{(x + 5) (x - 1)} \leq 0$

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 - x \cdot x - x \cdot - 1 + 3 (x - 1)}{(x + 5) (x - 1)} \leq 0$

Apply the distributive property.

$\frac{x^{2} + 8 x + 15 - x \cdot x - x \cdot - 1 + 3 x + 3 \cdot - 1}{(x + 5) (x - 1)} \leq 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$\frac{x^{2} + 8 x + 15 - (x \cdot x) - x \cdot - 1 + 3 x + 3 \cdot - 1}{(x + 5) (x - 1)} \leq 0$

Multiply $x$ by $x$ .

$\frac{x^{2} + 8 x + 15 - x^{2} - x \cdot - 1 + 3 x + 3 \cdot - 1}{(x + 5) (x - 1)} \leq 0$

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$\frac{x^{2} + 8 x + 15 - x^{2} + 1 x + 3 x + 3 \cdot - 1}{(x + 5) (x - 1)} \leq 0$

Multiply $x$ by $1$ .

$\frac{x^{2} + 8 x + 15 - x^{2} + x + 3 x + 3 \cdot - 1}{(x + 5) (x - 1)} \leq 0$

Multiply $3$ by $- 1$ .

$\frac{x^{2} + 8 x + 15 - x^{2} + x + 3 x - 3}{(x + 5) (x - 1)} \leq 0$

Add $x$ and $3 x$ .

$\frac{x^{2} + 8 x + 15 - x^{2} + 4 x - 3}{(x + 5) (x - 1)} \leq 0$

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

$\frac{8 x + 15 + 0 + 4 x - 3}{(x + 5) (x - 1)} \leq 0$

Add $8 x$ and $0$ .

$\frac{8 x + 15 + 4 x - 3}{(x + 5) (x - 1)} \leq 0$

Add $8 x$ and $4 x$ .

$\frac{12 x + 15 - 3}{(x + 5) (x - 1)} \leq 0$

Subtract $3$ from $15$ .

$\frac{12 x + 12}{(x + 5) (x - 1)} \leq 0$

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

Factor $12$ out of $12 x$ .

$\frac{12 (x) + 12}{(x + 5) (x - 1)} \leq 0$

Factor $12$ out of $12$ .

$\frac{12 (x) + 12 (1)}{(x + 5) (x - 1)} \leq 0$

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

$\frac{12 (x + 1)}{(x + 5) (x - 1)} \leq 0$

Step 3

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

$x + 1 = 0$

$x + 5 = 0$

$x - 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

Add $1$ to both sides of the equation.

$x = 1$

Step 7

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

$x = - 1$

$x = - 5$

$x = 1$

Step 8

Consolidate the solutions.

$x = - 1, - 5, 1$

Step 9

Find the domain of $\frac{12 (x + 1)}{(x + 5) (x - 1)}$ .

Set the denominator in $\frac{12 (x + 1)}{(x + 5) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(x + 5) (x - 1) = 0$

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 $0$ .

$x + 5 = 0$

$x - 1 = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x + 5) (x - 1) = 0$ true.

$x = - 5, 1$

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

$(- \infty, - 5) \cup (- 5, 1) \cup (1, \infty)$

Step 10

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 11

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$\frac{(- 8) + 3}{(- 8) - 1} \leq \frac{(- 8) - 3}{(- 8) + 5}$

The left side $0.\overline{5}$ is less than the right side $3.\overline{6}$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 1$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

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

The left side $0$ is greater than the right side $- 3$ , 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.

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

The left side $- 3$ is less than the right side $- 0.6$ , 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.

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

The left side $2.\overline{3}$ is greater than the right side $0.\overline{1}$ , which means that the given statement is false.

False

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

$x < - 5$ True

$- 5 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 5$ True

$- 5 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

Step 12

The solution consists of all of the true intervals.

$x < - 5$ or $- 1 \leq x < 1$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 5 or - 1 \leq x < 1$

Interval Notation:

$(- \infty, - 5) \cup [- 1, 1)$

Step 14

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

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