Solve the Inequality for x (x-1)/(x^2-6x-7)>0

$\frac{x - 1}{x^{2} - 6 x - 7} > 0$

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

$x - 1 = 0$

$x^{2} - 6 x - 7 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Factor $x^{2} - 6 x - 7$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Write the factored form using these integers.

$(x - 7) (x + 1) = 0$

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

$x - 7 = 0$

$x + 1 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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 $(x - 7) (x + 1) = 0$ true.

$x = 7, - 1$

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

$x = 1$

$x = 7, - 1$

Consolidate the solutions.

$x = 1, 7, - 1$

Find the domain of $\frac{x - 1}{x^{2} - 6 x - 7}$ .

Set the denominator in $\frac{x - 1}{x^{2} - 6 x - 7}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 6 x - 7 = 0$

Solve for $x$ .

Factor $x^{2} - 6 x - 7$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Write the factored form using these integers.

$(x - 7) (x + 1) = 0$

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

$x - 7 = 0$

$x + 1 = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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 $(x - 7) (x + 1) = 0$ true.

$x = 7, - 1$

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

$(- \infty, - 1) \cup (- 1, 7) \cup (7, \infty)$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 7$

$x > 7$

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 $x < - 1$ 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.

$\frac{(- 4) - 1}{{(- 4)}^{2} - 6 \cdot - 4 - 7} > 0$

The left side $- 0.\overline{15}$ is not greater than the right side $0$ , 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.

$\frac{(0) - 1}{{(0)}^{2} - 6 \cdot 0 - 7} > 0$

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

True

Test a value on the interval $1 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{(4) - 1}{{(4)}^{2} - 6 \cdot 4 - 7} > 0$

The left side $- 0.2$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 7$ to see if it makes the inequality true.

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

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

$\frac{(10) - 1}{{(10)}^{2} - 6 \cdot 10 - 7} > 0$

The left side $0.\overline{27}$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 7$ False

$x > 7$ True

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 7$ False

$x > 7$ True

The solution consists of all of the true intervals.

$- 1 < x < 1$ or $x > 7$

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 1 or x > 7$

Interval Notation:

$(- 1, 1) \cup (7, \infty)$

Solve the Inequality for x (x-1)/(x^2-6x-7)>0

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