Solve the Inequality for x (10-x)/(x-3)>0

$\frac{10 - x}{x - 3} > 0$

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

$10 - x = 0$

$x - 3 = 0$

Subtract $10$ from both sides of the equation.

$- x = - 10$

Divide each term in $- x = - 10$ by $- 1$ and simplify.

Divide each term in $- x = - 10$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 10}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 10}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 10}{- 1}$

Simplify the right side.

Divide $- 10$ by $- 1$ .

$x = 10$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 10$

$x = 3$

Consolidate the solutions.

$x = 10, 3$

Find the domain of $\frac{10 - x}{x - 3}$ .

Set the denominator in $\frac{10 - x}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

Use each root to create test intervals.

$x < 3$

$3 < x < 10$

$x > 10$

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.

$x = 0$

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

$\frac{10 - (0)}{(0) - 3} > 0$

The left side $- 3.\overline{3}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $3 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 10$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{10 - (6)}{(6) - 3} > 0$

The left side $1.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

$\frac{10 - (12)}{(12) - 3} > 0$

The left side $- 0.\overline{2}$ is not 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.

$x < 3$ False

$3 < x < 10$ True

$x > 10$ False

$x < 3$ False

$3 < x < 10$ True

$x > 10$ False

The solution consists of all of the true intervals.

$3 < x < 10$

The result can be shown in multiple forms.

Inequality Form:

$3 < x < 10$

Interval Notation:

$(3, 10)$

Solve the Inequality for x (10-x)/(x-3)>0

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