Solve the Inequality for x 7/(x+5)<=8/(x+6)

$\frac{7}{x + 5} \leq \frac{8}{x + 6}$

Step 1

Subtract $\frac{8}{x + 6}$ from both sides of the inequality.

$\frac{7}{x + 5} - \frac{8}{x + 6} \leq 0$

Step 2

Simplify $\frac{7}{x + 5} - \frac{8}{x + 6}$ .

To write $\frac{7}{x + 5}$ as a fraction with a common denominator, multiply by $\frac{x + 6}{x + 6}$ .

$\frac{7}{x + 5} \cdot \frac{x + 6}{x + 6} - \frac{8}{x + 6} \leq 0$

To write $- \frac{8}{x + 6}$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$\frac{7}{x + 5} \cdot \frac{x + 6}{x + 6} - \frac{8}{x + 6} \cdot \frac{x + 5}{x + 5} \leq 0$

Write each expression with a common denominator of $(x + 5) (x + 6)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{7}{x + 5}$ by $\frac{x + 6}{x + 6}$ .

$\frac{7 (x + 6)}{(x + 5) (x + 6)} - \frac{8}{x + 6} \cdot \frac{x + 5}{x + 5} \leq 0$

Multiply $\frac{8}{x + 6}$ by $\frac{x + 5}{x + 5}$ .

$\frac{7 (x + 6)}{(x + 5) (x + 6)} - \frac{8 (x + 5)}{(x + 6) (x + 5)} \leq 0$

Reorder the factors of $(x + 6) (x + 5)$ .

$\frac{7 (x + 6)}{(x + 5) (x + 6)} - \frac{8 (x + 5)}{(x + 5) (x + 6)} \leq 0$

Combine the numerators over the common denominator.

$\frac{7 (x + 6) - 8 (x + 5)}{(x + 5) (x + 6)} \leq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{7 x + 7 \cdot 6 - 8 (x + 5)}{(x + 5) (x + 6)} \leq 0$

Multiply $7$ by $6$ .

$\frac{7 x + 42 - 8 (x + 5)}{(x + 5) (x + 6)} \leq 0$

Apply the distributive property.

$\frac{7 x + 42 - 8 x - 8 \cdot 5}{(x + 5) (x + 6)} \leq 0$

Multiply $- 8$ by $5$ .

$\frac{7 x + 42 - 8 x - 40}{(x + 5) (x + 6)} \leq 0$

Subtract $8 x$ from $7 x$ .

$\frac{- x + 42 - 40}{(x + 5) (x + 6)} \leq 0$

Subtract $40$ from $42$ .

$\frac{- x + 2}{(x + 5) (x + 6)} \leq 0$

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

$\frac{- (x) + 2}{(x + 5) (x + 6)} \leq 0$

Rewrite $2$ as $- 1 (- 2)$ .

$\frac{- (x) - 1 (- 2)}{(x + 5) (x + 6)} \leq 0$

Factor $- 1$ out of $- (x) - 1 (- 2)$ .

$\frac{- (x - 2)}{(x + 5) (x + 6)} \leq 0$

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

$\frac{- 1 (x - 2)}{(x + 5) (x + 6)} \leq 0$

Move the negative in front of the fraction.

$- \frac{x - 2}{(x + 5) (x + 6)} \leq 0$

Step 3

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

$x - 2 = 0$

$x + 5 = 0$

$x + 6 = 0$

Step 4

Add $2$ to both sides of the equation.

$x = 2$

Step 5

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 7

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

$x = 2$

$x = - 5$

$x = - 6$

Step 8

Consolidate the solutions.

$x = 2, - 5, - 6$

Step 9

Find the domain of $- \frac{x - 2}{(x + 5) (x + 6)}$ .

Set the denominator in $\frac{x - 2}{(x + 5) (x + 6)}$ equal to $0$ to find where the expression is undefined.

$(x + 5) (x + 6) = 0$

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 $0$ .

$x + 5 = 0$

$x + 6 = 0$

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 $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

The final solution is all the values that make $(x + 5) (x + 6) = 0$ true.

$x = - 5, - 6$

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

$(- \infty, - 6) \cup (- 6, - 5) \cup (- 5, \infty)$

Step 10

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 5$

$- 5 < x < 2$

$x > 2$

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 $x < - 6$ to see if it makes the inequality true.

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

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

$\frac{7}{(- 8) + 5} \leq \frac{8}{(- 8) + 6}$

The left side $- 2.\overline{3}$ is greater than the right side $- 4$ , which means that the given statement is false.

False

Test a value on the interval $- 6 < x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < x < - 5$ and see if this value makes the original inequality true.

$x = - 5.5$

Replace $x$ with $- 5.5$ in the original inequality.

$\frac{7}{(- 5.5) + 5} \leq \frac{8}{(- 5.5) + 6}$

The left side $- 14$ is less than the right side $16$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{7}{(0) + 5} \leq \frac{8}{(0) + 6}$

The left side $1.4$ is greater than the right side $1.\overline{3}$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{7}{(4) + 5} \leq \frac{8}{(4) + 6}$

The left side $0.\overline{7}$ is less than the right side $0.8$ , which means that the given statement is always true.

True

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

$x < - 6$ False

$- 6 < x < - 5$ True

$- 5 < x < 2$ False

$x > 2$ True

$x < - 6$ False

$- 6 < x < - 5$ True

$- 5 < x < 2$ False

$x > 2$ True

Step 12

The solution consists of all of the true intervals.

$- 6 < x < - 5$ or $x \geq 2$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 6 < x < - 5 or x \geq 2$

Interval Notation:

$(- 6, - 5) \cup [2, \infty)$

Step 14

Solve the Inequality for x 7/(x+5)<=8/(x+6)

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