Solve the Inequality for x (x+3)/(x+8)<(x-1)/(x-3)

$\frac{x + 3}{x + 8} < \frac{x - 1}{x - 3}$

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

$\frac{x + 3}{x + 8} - \frac{x - 1}{x - 3} < 0$

Simplify $\frac{x + 3}{x + 8} - \frac{x - 1}{x - 3}$ .

To write $\frac{x + 3}{x + 8}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{x + 3}{x + 8} \cdot \frac{x - 3}{x - 3} - \frac{x - 1}{x - 3} < 0$

To write $- \frac{x - 1}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x + 8}{x + 8}$ .

$\frac{x + 3}{x + 8} \cdot \frac{x - 3}{x - 3} - \frac{x - 1}{x - 3} \cdot \frac{x + 8}{x + 8} < 0$

Write each expression with a common denominator of $(x + 8) (x - 3)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x + 3}{x + 8}$ by $\frac{x - 3}{x - 3}$ .

$\frac{(x + 3) (x - 3)}{(x + 8) (x - 3)} - \frac{x - 1}{x - 3} \cdot \frac{x + 8}{x + 8} < 0$

Multiply $\frac{x - 1}{x - 3}$ by $\frac{x + 8}{x + 8}$ .

$\frac{(x + 3) (x - 3)}{(x + 8) (x - 3)} - \frac{(x - 1) (x + 8)}{(x - 3) (x + 8)} < 0$

Reorder the factors of $(x - 3) (x + 8)$ .

$\frac{(x + 3) (x - 3)}{(x + 8) (x - 3)} - \frac{(x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Combine the numerators over the common denominator.

$\frac{(x + 3) (x - 3) - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Simplify the numerator.

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

Apply the distributive property.

$\frac{x (x - 3) + 3 (x - 3) - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 3 + 3 (x - 3) - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3 - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Combine the opposite terms in $x \cdot x + x \cdot - 3 + 3 x + 3 \cdot - 3$ .

Reorder the factors in the terms $x \cdot - 3$ and $3 x$ .

$\frac{x \cdot x - 3 x + 3 x + 3 \cdot - 3 - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

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

$\frac{x \cdot x + 0 + 3 \cdot - 3 - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Add $x \cdot x$ and $0$ .

$\frac{x \cdot x + 3 \cdot - 3 - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Simplify each term.

Multiply $x$ by $x$ .

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

Multiply $3$ by $- 3$ .

$\frac{x^{2} - 9 - (x - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Apply the distributive property.

$\frac{x^{2} - 9 + (- x - - 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Multiply $- 1$ by $- 1$ .

$\frac{x^{2} - 9 + (- x + 1) (x + 8)}{(x + 8) (x - 3)} < 0$

Expand $(- x + 1) (x + 8)$ using the FOIL Method.

Apply the distributive property.

$\frac{x^{2} - 9 - x (x + 8) + 1 (x + 8)}{(x + 8) (x - 3)} < 0$

Apply the distributive property.

$\frac{x^{2} - 9 - x \cdot x - x \cdot 8 + 1 (x + 8)}{(x + 8) (x - 3)} < 0$

Apply the distributive property.

$\frac{x^{2} - 9 - x \cdot x - x \cdot 8 + 1 x + 1 \cdot 8}{(x + 8) (x - 3)} < 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$\frac{x^{2} - 9 - (x \cdot x) - x \cdot 8 + 1 x + 1 \cdot 8}{(x + 8) (x - 3)} < 0$

Multiply $x$ by $x$ .

$\frac{x^{2} - 9 - x^{2} - x \cdot 8 + 1 x + 1 \cdot 8}{(x + 8) (x - 3)} < 0$

Multiply $8$ by $- 1$ .

$\frac{x^{2} - 9 - x^{2} - 8 x + 1 x + 1 \cdot 8}{(x + 8) (x - 3)} < 0$

Multiply $x$ by $1$ .

$\frac{x^{2} - 9 - x^{2} - 8 x + x + 1 \cdot 8}{(x + 8) (x - 3)} < 0$

Multiply $8$ by $1$ .

$\frac{x^{2} - 9 - x^{2} - 8 x + x + 8}{(x + 8) (x - 3)} < 0$

Add $- 8 x$ and $x$ .

$\frac{x^{2} - 9 - x^{2} - 7 x + 8}{(x + 8) (x - 3)} < 0$

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

$\frac{0 - 9 - 7 x + 8}{(x + 8) (x - 3)} < 0$

Subtract $9$ from $0$ .

$\frac{- 9 - 7 x + 8}{(x + 8) (x - 3)} < 0$

Add $- 9$ and $8$ .

$\frac{- 7 x - 1}{(x + 8) (x - 3)} < 0$

Factor $- 1$ out of $- 7 x$ .

$\frac{- (7 x) - 1}{(x + 8) (x - 3)} < 0$

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

$\frac{- (7 x) - 1 (1)}{(x + 8) (x - 3)} < 0$

Factor $- 1$ out of $- (7 x) - 1 (1)$ .

$\frac{- (7 x + 1)}{(x + 8) (x - 3)} < 0$

Rewrite $- (7 x + 1)$ as $- 1 (7 x + 1)$ .

$\frac{- 1 (7 x + 1)}{(x + 8) (x - 3)} < 0$

Move the negative in front of the fraction.

$- \frac{7 x + 1}{(x + 8) (x - 3)} < 0$

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

$7 x + 1 = 0$

$x + 8 = 0$

$x - 3 = 0$

Subtract $1$ from both sides of the equation.

$7 x = - 1$

Divide each term in $7 x = - 1$ by $7$ and simplify.

Divide each term in $7 x = - 1$ by $7$ .

$\frac{7 x}{7} = \frac{- 1}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 1}{7}$

Divide $x$ by $1$ .

$x = \frac{- 1}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{7}$

Subtract $8$ from both sides of the equation.

$x = - 8$

Add $3$ to 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.

$x = - \frac{1}{7}$

$x = - 8$

$x = 3$

Consolidate the solutions.

$x = - \frac{1}{7}, - 8, 3$

Find the domain of $- \frac{7 x + 1}{(x + 8) (x - 3)}$ .

Set the denominator in $\frac{7 x + 1}{(x + 8) (x - 3)}$ equal to $0$ to find where the expression is undefined.

$(x + 8) (x - 3) = 0$

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 $0$ .

$x + 8 = 0$

$x - 3 = 0$

Set $x + 8$ equal to $0$ and solve for $x$ .

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Subtract $8$ from both sides of the equation.

$x = - 8$

Set $x - 3$ equal to $0$ and solve for $x$ .

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x + 8) (x - 3) = 0$ true.

$x = - 8, 3$

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

$(- \infty, - 8) \cup (- 8, 3) \cup (3, \infty)$

Use each root to create test intervals.

$x < - 8$

$- 8 < x < - \frac{1}{7}$

$- \frac{1}{7} < x < 3$

$x > 3$

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 $x < - 8$ to see if it makes the inequality true.

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$\frac{(- 10) + 3}{(- 10) + 8} < \frac{(- 10) - 1}{(- 10) - 3}$

The left side $3.5$ is not less than the right side $0.\overline{846153}$ , which means that the given statement is false.

False

Test a value on the interval $- 8 < x < - \frac{1}{7}$ to see if it makes the inequality true.

Choose a value on the interval $- 8 < x < - \frac{1}{7}$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{(- 4) + 3}{(- 4) + 8} < \frac{(- 4) - 1}{(- 4) - 3}$

The left side $- 0.25$ is less than the right side $0.\overline{714285}$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{1}{7} < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{7} < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 3}{(0) + 8} < \frac{(0) - 1}{(0) - 3}$

The left side $0.375$ is not less than the right side $0.\overline{3}$ , which means that the given statement is false.

False

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.

$\frac{(6) + 3}{(6) + 8} < \frac{(6) - 1}{(6) - 3}$

The left side $0.6\overline{428571}$ is less than the right side $1.\overline{6}$ , which means that the given statement is always true.

True

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

$x < - 8$ False

$- 8 < x < - \frac{1}{7}$ True

$- \frac{1}{7} < x < 3$ False

$x > 3$ True

$x < - 8$ False

$- 8 < x < - \frac{1}{7}$ True

$- \frac{1}{7} < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$- 8 < x < - \frac{1}{7}$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$- 8 < x < - \frac{1}{7} or x > 3$

Interval Notation:

$(- 8, - \frac{1}{7}) \cup (3, \infty)$

Solve the Inequality for x (x+3)/(x+8)<(x-1)/(x-3)

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