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

$\frac{2 x - 1}{3 x - 2} \leq 1$

Subtract $1$ from both sides of the inequality.

$\frac{2 x - 1}{3 x - 2} - 1 \leq 0$

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3 x - 2}{3 x - 2}$ .

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

Combine $- 1$ and $\frac{3 x - 2}{3 x - 2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $3$ by $- 1$ .

$\frac{2 x - 1 - 3 x - - 2}{3 x - 2} \leq 0$

Multiply $- 1$ by $- 2$ .

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

Subtract $3 x$ from $2 x$ .

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

Add $- 1$ and $2$ .

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

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

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

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

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

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

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

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

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

Move the negative in front of the fraction.

$- \frac{x - 1}{3 x - 2} \leq 0$

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

$x - 1 = 0$

$3 x - 2 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

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

$x = 1$

$x = \frac{2}{3}$

Consolidate the solutions.

$x = 1, \frac{2}{3}$

Find the domain of $- \frac{x - 1}{3 x - 2}$ .

Set the denominator in $\frac{x - 1}{3 x - 2}$ equal to $0$ to find where the expression is undefined.

$3 x - 2 = 0$

Solve for $x$ .

Add $2$ to both sides of the equation.

$3 x = 2$

Divide each term in $3 x = 2$ by $3$ and simplify.

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Divide $x$ by $1$ .

$x = \frac{2}{3}$

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

$(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty)$

Use each root to create test intervals.

$x < \frac{2}{3}$

$\frac{2}{3} < x < 1$

$x > 1$

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 $x < \frac{2}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{2}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $0.5$ is less than the right side $1$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{2}{3} < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $\frac{2}{3} < x < 1$ and see if this value makes the original inequality true.

$x = 0.83$

Replace $x$ with $0.83$ in the original inequality.

$\frac{2 (0.83) - 1}{3 (0.83) - 2} \leq 1$

The left side $1.34693877$ is greater than the right side $1$ , 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.

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

The left side $0.7$ is less than the right side $1$ , which means that the given statement is always true.

True

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

$x < \frac{2}{3}$ True

$\frac{2}{3} < x < 1$ False

$x > 1$ True

$x < \frac{2}{3}$ True

$\frac{2}{3} < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$x < \frac{2}{3}$ or $x \geq 1$

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{2}{3} or x \geq 1$

Interval Notation:

$(- \infty, \frac{2}{3}) \cup [1, \infty)$

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

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