Solve the Inequality for x 3/(x-1)-4/x>=1

$\frac{3}{x - 1} - \frac{4}{x} \geq 1$

Subtract $1$ from both sides of the inequality.

$\frac{3}{x - 1} - \frac{4}{x} - 1 \geq 0$

Simplify $\frac{3}{x - 1} - \frac{4}{x} - 1$ .

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

$\frac{3}{x - 1} \cdot \frac{x}{x} - \frac{4}{x} - 1 \geq 0$

To write $- \frac{4}{x}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{3}{x - 1} \cdot \frac{x}{x} - \frac{4}{x} \cdot \frac{x - 1}{x - 1} - 1 \geq 0$

Write each expression with a common denominator of $(x - 1) x$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{x - 1}$ by $\frac{x}{x}$ .

$\frac{3 x}{(x - 1) x} - \frac{4}{x} \cdot \frac{x - 1}{x - 1} - 1 \geq 0$

Multiply $\frac{4}{x}$ by $\frac{x - 1}{x - 1}$ .

$\frac{3 x}{(x - 1) x} - \frac{4 (x - 1)}{x (x - 1)} - 1 \geq 0$

Reorder the factors of $(x - 1) x$ .

$\frac{3 x}{x (x - 1)} - \frac{4 (x - 1)}{x (x - 1)} - 1 \geq 0$

Combine the numerators over the common denominator.

$\frac{3 x - 4 (x - 1)}{x (x - 1)} - 1 \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{3 x - 4 x - 4 \cdot - 1}{x (x - 1)} - 1 \geq 0$

Multiply $- 4$ by $- 1$ .

$\frac{3 x - 4 x + 4}{x (x - 1)} - 1 \geq 0$

Subtract $4 x$ from $3 x$ .

$\frac{- x + 4}{x (x - 1)} - 1 \geq 0$

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

$\frac{- x + 4}{x (x - 1)} - 1 \cdot \frac{x (x - 1)}{x (x - 1)} \geq 0$

Combine $- 1$ and $\frac{x (x - 1)}{x (x - 1)}$ .

$\frac{- x + 4}{x (x - 1)} + \frac{- (x (x - 1))}{x (x - 1)} \geq 0$

Combine the numerators over the common denominator.

$\frac{- x + 4 - (x (x - 1))}{x (x - 1)} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{- x + 4 - x \cdot x - x \cdot - 1}{x (x - 1)} \geq 0$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\frac{- x + 4 - (x \cdot x) - x \cdot - 1}{x (x - 1)} \geq 0$

Multiply $x$ by $x$ .

$\frac{- x + 4 - x^{2} - x \cdot - 1}{x (x - 1)} \geq 0$

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$\frac{- x + 4 - x^{2} + 1 x}{x (x - 1)} \geq 0$

Multiply $x$ by $1$ .

$\frac{- x + 4 - x^{2} + x}{x (x - 1)} \geq 0$

Add $- x$ and $x$ .

$\frac{4 - x^{2} + 0}{x (x - 1)} \geq 0$

Add $4 - x^{2}$ and $0$ .

$\frac{4 - x^{2}}{x (x - 1)} \geq 0$

Rewrite $4 - x^{2}$ in a factored form.

Rewrite $4$ as $2^{2}$ .

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

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

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

$x = 0$

$2 + x = 0$

$2 - x = 0$

$x - 1 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Subtract $2$ from both sides of the equation.

$- x = - 2$

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$x = 2$

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.

$x = 0$

$x = - 2$

$x = 2$

$x = 1$

Consolidate the solutions.

$x = 0, - 2, 2, 1$

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

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

$x (x - 1) = 0$

Solve for $x$ .

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

$x = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $x (x - 1) = 0$ true.

$x = 0, 1$

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

$(- \infty, 0) \cup (0, 1) \cup (1, \infty)$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$0 < x < 1$

$1 < x < 2$

$x > 2$

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 $x < - 2$ 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.

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

The left side $0.4$ is less than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

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

The left side $2.5$ is greater than the right side $1$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Replace $x$ with $0.5$ in the original inequality.

$\frac{3}{(0.5) - 1} - \frac{4}{0.5} \geq 1$

The left side $- 14$ is less than the right side $1$ , 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.

$\frac{3}{(1.5) - 1} - \frac{4}{1.5} \geq 1$

The left side $3.\overline{3}$ is greater than the right side $1$ , 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.

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

The left side $0$ is less than the right side $1$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < 0$ True

$0 < x < 1$ False

$1 < x < 2$ True

$x > 2$ False

$x < - 2$ False

$- 2 < x < 0$ True

$0 < x < 1$ False

$1 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$- 2 \leq x < 0$ or $1 < x \leq 2$

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x < 0 or 1 < x \leq 2$

Interval Notation:

$[- 2, 0) \cup (1, 2]$

Solve the Inequality for x 3/(x-1)-4/x>=1

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