Solve the Inequality for x 8/(x-3)>6/(x-1)

$\frac{8}{x - 3} > \frac{6}{x - 1}$

Subtract $\frac{6}{x - 1}$ from both sides of the inequality.

$\frac{8}{x - 3} - \frac{6}{x - 1} > 0$

Simplify $\frac{8}{x - 3} - \frac{6}{x - 1}$ .

To write $\frac{8}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{8}{x - 3} \cdot \frac{x - 1}{x - 1} - \frac{6}{x - 1} > 0$

To write $- \frac{6}{x - 1}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{8}{x - 3} \cdot \frac{x - 1}{x - 1} - \frac{6}{x - 1} \cdot \frac{x - 3}{x - 3} > 0$

Write each expression with a common denominator of $(x - 3) (x - 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{8}{x - 3}$ by $\frac{x - 1}{x - 1}$ .

$\frac{8 (x - 1)}{(x - 3) (x - 1)} - \frac{6}{x - 1} \cdot \frac{x - 3}{x - 3} > 0$

Multiply $\frac{6}{x - 1}$ by $\frac{x - 3}{x - 3}$ .

$\frac{8 (x - 1)}{(x - 3) (x - 1)} - \frac{6 (x - 3)}{(x - 1) (x - 3)} > 0$

Reorder the factors of $(x - 3) (x - 1)$ .

$\frac{8 (x - 1)}{(x - 1) (x - 3)} - \frac{6 (x - 3)}{(x - 1) (x - 3)} > 0$

Combine the numerators over the common denominator.

$\frac{8 (x - 1) - 6 (x - 3)}{(x - 1) (x - 3)} > 0$

Simplify the numerator.

Factor $2$ out of $8 (x - 1) - 6 (x - 3)$ .

Factor $2$ out of $8 (x - 1)$ .

$\frac{2 (4 (x - 1)) - 6 (x - 3)}{(x - 1) (x - 3)} > 0$

Factor $2$ out of $- 6 (x - 3)$ .

$\frac{2 (4 (x - 1)) + 2 (- 3 (x - 3))}{(x - 1) (x - 3)} > 0$

Factor $2$ out of $2 (4 (x - 1)) + 2 (- 3 (x - 3))$ .

$\frac{2 (4 (x - 1) - 3 (x - 3))}{(x - 1) (x - 3)} > 0$

Apply the distributive property.

$\frac{2 (4 x + 4 \cdot - 1 - 3 (x - 3))}{(x - 1) (x - 3)} > 0$

Multiply $4$ by $- 1$ .

$\frac{2 (4 x - 4 - 3 (x - 3))}{(x - 1) (x - 3)} > 0$

Apply the distributive property.

$\frac{2 (4 x - 4 - 3 x - 3 \cdot - 3)}{(x - 1) (x - 3)} > 0$

Multiply $- 3$ by $- 3$ .

$\frac{2 (4 x - 4 - 3 x + 9)}{(x - 1) (x - 3)} > 0$

Subtract $3 x$ from $4 x$ .

$\frac{2 (x - 4 + 9)}{(x - 1) (x - 3)} > 0$

Add $- 4$ and $9$ .

$\frac{2 (x + 5)}{(x - 1) (x - 3)} > 0$

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

$x + 5 = 0$

$x - 1 = 0$

$x - 3 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Add $1$ to both sides of the equation.

$x = 1$

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.

$x = - 5$

$x = 1$

$x = 3$

Consolidate the solutions.

$x = - 5, 1, 3$

Find the domain of $\frac{2 (x + 5)}{(x - 1) (x - 3)}$ .

Set the denominator in $\frac{2 (x + 5)}{(x - 1) (x - 3)}$ equal to $0$ to find where the expression is undefined.

$(x - 1) (x - 3) = 0$

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 $0$ .

$x - 1 = 0$

$x - 3 = 0$

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.

$x = 1$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x - 1) (x - 3) = 0$ true.

$x = 1, 3$

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

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

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$1 < x < 3$

$x > 3$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$\frac{8}{(- 8) - 3} > \frac{6}{(- 8) - 1}$

The left side $- 0.\overline{72}$ is not greater than the right side $- 0.\overline{6}$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{8}{(0) - 3} > \frac{6}{(0) - 1}$

The left side $- 2.\overline{6}$ is greater than the right side $- 6$ , which means that the given statement is always true.

True

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

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

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\frac{8}{(2) - 3} > \frac{6}{(2) - 1}$

The left side $- 8$ is not greater than the right side $6$ , which means that the given statement is false.

False

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.

$\frac{8}{(6) - 3} > \frac{6}{(6) - 1}$

The left side $2.\overline{6}$ is greater than the right side $1.2$ , which means that the given statement is always true.

True

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

$x < - 5$ False

$- 5 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

$x < - 5$ False

$- 5 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$- 5 < x < 1$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x < 1 or x > 3$

Interval Notation:

$(- 5, 1) \cup (3, \infty)$

Solve the Inequality for x 8/(x-3)>6/(x-1)

Related Questions

Support LCMGCF.com by disabling your adblocker.