Solve the Inequality for x (2-5x)/(x-2)>=3

$\frac{2 - 5 x}{x - 2} \geq 3$

Subtract $3$ from both sides of the inequality.

$\frac{2 - 5 x}{x - 2} - 3 \geq 0$

Simplify $\frac{2 - 5 x}{x - 2} - 3$ .

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{2 - 5 x}{x - 2} - 3 \cdot \frac{x - 2}{x - 2} \geq 0$

Combine $- 3$ and $\frac{x - 2}{x - 2}$ .

$\frac{2 - 5 x}{x - 2} + \frac{- 3 (x - 2)}{x - 2} \geq 0$

Combine the numerators over the common denominator.

$\frac{2 - 5 x - 3 (x - 2)}{x - 2} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{2 - 5 x - 3 x - 3 \cdot - 2}{x - 2} \geq 0$

Multiply $- 3$ by $- 2$ .

$\frac{2 - 5 x - 3 x + 6}{x - 2} \geq 0$

Add $2$ and $6$ .

$\frac{- 5 x - 3 x + 8}{x - 2} \geq 0$

Subtract $3 x$ from $- 5 x$ .

$\frac{- 8 x + 8}{x - 2} \geq 0$

Factor $8$ out of $- 8 x + 8$ .

Factor $8$ out of $- 8 x$ .

$\frac{8 (- x) + 8}{x - 2} \geq 0$

Factor $8$ out of $8$ .

$\frac{8 (- x) + 8 (1)}{x - 2} \geq 0$

Factor $8$ out of $8 (- x) + 8 (1)$ .

$\frac{8 (- x + 1)}{x - 2} \geq 0$

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

$\frac{8 (- (x) + 1)}{x - 2} \geq 0$

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

$\frac{8 (- (x) - 1 (- 1))}{x - 2} \geq 0$

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

$\frac{8 (- (x - 1))}{x - 2} \geq 0$

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

$\frac{8 (- 1 (x - 1))}{x - 2} \geq 0$

Move the negative in front of the fraction.

$- \frac{8 (x - 1)}{x - 2} \geq 0$

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

$x - 1 = 0$

$x - 2 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Add $2$ to both sides of the equation.

$x = 2$

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

$x = 1$

$x = 2$

Consolidate the solutions.

$x = 1, 2$

Find the domain of $- \frac{8 (x - 1)}{x - 2}$ .

Set the denominator in $\frac{8 (x - 1)}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

$(- \infty, 2) \cup (2, \infty)$

Use each root to create test intervals.

$x < 1$

$1 < x < 2$

$x > 2$

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.

$x = 0$

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

$\frac{2 - 5 \cdot 0}{(0) - 2} \geq 3$

The left side $- 1$ is less than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Replace $x$ with $1.5$ in the original inequality.

$\frac{2 - 5 \cdot 1.5}{(1.5) - 2} \geq 3$

The left side $11$ is greater than the right side $3$ , which means that the given statement is always true.

True

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.

$\frac{2 - 5 \cdot 4}{(4) - 2} \geq 3$

The left side $- 9$ is less than the right side $3$ , which means that the given statement is false.

False

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

$x < 1$ False

$1 < x < 2$ True

$x > 2$ False

$x < 1$ False

$1 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$1 \leq x < 2$

The result can be shown in multiple forms.

Inequality Form:

$1 \leq x < 2$

Interval Notation:

$[1, 2)$

Solve the Inequality for x (2-5x)/(x-2)>=3

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