Convert to Interval Notation 4/(c-2)-3/c<0

$\frac{4}{c - 2} - \frac{3}{c} < 0$

Simplify $\frac{4}{c - 2} - \frac{3}{c}$ .

To write $\frac{4}{c - 2}$ as a fraction with a common denominator, multiply by $\frac{c}{c}$ .

$\frac{4}{c - 2} \cdot \frac{c}{c} - \frac{3}{c} < 0$

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

$\frac{4}{c - 2} \cdot \frac{c}{c} - \frac{3}{c} \cdot \frac{c - 2}{c - 2} < 0$

Write each expression with a common denominator of $(c - 2) c$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{c - 2}$ by $\frac{c}{c}$ .

$\frac{4 c}{(c - 2) c} - \frac{3}{c} \cdot \frac{c - 2}{c - 2} < 0$

Multiply $\frac{3}{c}$ by $\frac{c - 2}{c - 2}$ .

$\frac{4 c}{(c - 2) c} - \frac{3 (c - 2)}{c (c - 2)} < 0$

Reorder the factors of $(c - 2) c$ .

$\frac{4 c}{c (c - 2)} - \frac{3 (c - 2)}{c (c - 2)} < 0$

Combine the numerators over the common denominator.

$\frac{4 c - 3 (c - 2)}{c (c - 2)} < 0$

Simplify the numerator.

Apply the distributive property.

$\frac{4 c - 3 c - 3 \cdot - 2}{c (c - 2)} < 0$

Multiply $- 3$ by $- 2$ .

$\frac{4 c - 3 c + 6}{c (c - 2)} < 0$

Subtract $3 c$ from $4 c$ .

$\frac{c + 6}{c (c - 2)} < 0$

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

$c = 0$

$c + 6 = 0$

$c - 2 = 0$

Subtract $6$ from both sides of the equation.

$c = - 6$

Add $2$ to both sides of the equation.

$c = 2$

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

$c = 0$

$c = - 6$

$c = 2$

Consolidate the solutions.

$c = 0, - 6, 2$

Find the domain of $\frac{4}{c - 2} - \frac{3}{c}$ .

Set the denominator in $\frac{4}{c - 2}$ equal to $0$ to find where the expression is undefined.

$c - 2 = 0$

Add $2$ to both sides of the equation.

$c = 2$

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

$c = 0$

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

$(- \infty, 0) \cup (0, 2) \cup (2, \infty)$

Use each root to create test intervals.

$c < - 6$

$- 6 < c < 0$

$0 < c < 2$

$c > 2$

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 $c < - 6$ to see if it makes the inequality true.

Choose a value on the interval $c < - 6$ and see if this value makes the original inequality true.

$c = - 8$

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

$\frac{4}{(- 8) - 2} - \frac{3}{- 8} < 0$

The left side $- 0.025$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 6 < c < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 6 < c < 0$ and see if this value makes the original inequality true.

$c = - 3$

Replace $c$ with $- 3$ in the original inequality.

$\frac{4}{(- 3) - 2} - \frac{3}{- 3} < 0$

The left side $0.2$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

$c = 1$

Replace $c$ with $1$ in the original inequality.

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

The left side $- 7$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $c > 2$ to see if it makes the inequality true.

Choose a value on the interval $c > 2$ and see if this value makes the original inequality true.

$c = 4$

Replace $c$ with $4$ in the original inequality.

$\frac{4}{(4) - 2} - \frac{3}{4} < 0$

The left side $1.25$ is not less 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.

$c < - 6$ True

$- 6 < c < 0$ False

$0 < c < 2$ True

$c > 2$ False

$c < - 6$ True

$- 6 < c < 0$ False

$0 < c < 2$ True

$c > 2$ False

The solution consists of all of the true intervals.

$c < - 6$ or $0 < c < 2$

Convert the inequality to interval notation.

$(- \infty, - 6) \cup (0, 2)$

Convert to Interval Notation 4/(c-2)-3/c<0

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