Solve the Inequality for x (3x+1)/(x-7)<=0

$\frac{3 x + 1}{x - 7} \leq 0$

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

$3 x + 1 = 0$

$x - 7 = 0$

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Add $7$ to both sides of the equation.

$x = 7$

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

$x = - \frac{1}{3}$

$x = 7$

Consolidate the solutions.

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

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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

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

Use each root to create test intervals.

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

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

$x > 7$

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 $x < - \frac{1}{3}$ to see if it makes the inequality true.

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

$x = - 3$

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

$\frac{3 (- 3) + 1}{(- 3) - 7} \leq 0$

The left side $0.8$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$\frac{3 (0) + 1}{(0) - 7} \leq 0$

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

True

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.

$\frac{3 (10) + 1}{(10) - 7} \leq 0$

The left side $10.\overline{3}$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

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

$x > 7$ False

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

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

$x > 7$ False

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

$[- \frac{1}{3}, 7)$

Solve the Inequality for x (3x+1)/(x-7)<=0

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