Solve the Inequality for x (3x+1)/(x-1)>=2

$\frac{3 x + 1}{x - 1} \geq 2$

Subtract $2$ from both sides of the inequality.

$\frac{3 x + 1}{x - 1} - 2 \geq 0$

Simplify $\frac{3 x + 1}{x - 1} - 2$ .

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{3 x + 1}{x - 1} - 2 \cdot \frac{x - 1}{x - 1} \geq 0$

Combine $- 2$ and $\frac{x - 1}{x - 1}$ .

$\frac{3 x + 1}{x - 1} + \frac{- 2 (x - 1)}{x - 1} \geq 0$

Combine the numerators over the common denominator.

$\frac{3 x + 1 - 2 (x - 1)}{x - 1} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 x + 1 - 2 x - 2 \cdot - 1}{x - 1} \geq 0$

Multiply $- 2$ by $- 1$ .

$\frac{3 x + 1 - 2 x + 2}{x - 1} \geq 0$

Subtract $2 x$ from $3 x$ .

$\frac{x + 1 + 2}{x - 1} \geq 0$

Add $1$ and $2$ .

$\frac{x + 3}{x - 1} \geq 0$

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

$x + 3 = 0$

$x - 1 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = - 3$

$x = 1$

Consolidate the solutions.

$x = - 3, 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 1$

$x > 1$

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 $x < - 3$ 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.

$\frac{3 (- 6) + 1}{(- 6) - 1} \geq 2$

The left side $2.\overline{428571}$ is greater than the right side $2$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$\frac{3 (0) + 1}{(0) - 1} \geq 2$

The left side $- 1$ is less than the right side $2$ , which means that the given statement is false.

False

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.

$\frac{3 (4) + 1}{(4) - 1} \geq 2$

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

True

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

$x < - 3$ True

$- 3 < x < 1$ False

$x > 1$ True

$x < - 3$ True

$- 3 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$x \leq - 3$ or $x > 1$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x > 1$

Interval Notation:

$(- \infty, - 3] \cup (1, \infty)$

Solve the Inequality for x (3x+1)/(x-1)>=2

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