Solve for x x/3-1/(x-2)<(x+1)/4

$\frac{x}{3} - \frac{1}{x - 2} < \frac{x + 1}{4}$

Step 1

Subtract $\frac{x + 1}{4}$ from both sides of the inequality.

$\frac{x}{3} - \frac{1}{x - 2} - \frac{x + 1}{4} < 0$

Step 2

Simplify $\frac{x}{3} - \frac{1}{x - 2} - \frac{x + 1}{4}$ .

Find the common denominator.

Multiply $\frac{x}{3}$ by $\frac{4 (x - 2)}{4 (x - 2)}$ .

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

Multiply $\frac{x}{3}$ by $\frac{4 (x - 2)}{4 (x - 2)}$ .

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

Multiply $\frac{1}{x - 2}$ by $\frac{12}{12}$ .

$\frac{x (4 (x - 2))}{3 (4 (x - 2))} - (\frac{1}{x - 2} \cdot \frac{12}{12}) - \frac{x + 1}{4} < 0$

Multiply $\frac{1}{x - 2}$ by $\frac{12}{12}$ .

$\frac{x (4 (x - 2))}{3 (4 (x - 2))} - \frac{12}{(x - 2) \cdot 12} - \frac{x + 1}{4} < 0$

Multiply $\frac{x + 1}{4}$ by $\frac{3 (x - 2)}{3 (x - 2)}$ .

$\frac{x (4 (x - 2))}{3 (4 (x - 2))} - \frac{12}{(x - 2) \cdot 12} - (\frac{x + 1}{4} \cdot \frac{3 (x - 2)}{3 (x - 2)}) < 0$

Multiply $\frac{x + 1}{4}$ by $\frac{3 (x - 2)}{3 (x - 2)}$ .

$\frac{x (4 (x - 2))}{3 (4 (x - 2))} - \frac{12}{(x - 2) \cdot 12} - \frac{(x + 1) (3 (x - 2))}{4 (3 (x - 2))} < 0$

Reorder the factors of $3 (4 (x - 2))$ .

$\frac{x (4 (x - 2))}{3 \cdot 4 (x - 2)} - \frac{12}{(x - 2) \cdot 12} - \frac{(x + 1) (3 (x - 2))}{4 (3 (x - 2))} < 0$

Multiply $3$ by $4$ .

$\frac{x (4 (x - 2))}{12 (x - 2)} - \frac{12}{(x - 2) \cdot 12} - \frac{(x + 1) (3 (x - 2))}{4 (3 (x - 2))} < 0$

Reorder the factors of $(x - 2) \cdot 12$ .

$\frac{x (4 (x - 2))}{12 (x - 2)} - \frac{12}{12 (x - 2)} - \frac{(x + 1) (3 (x - 2))}{4 (3 (x - 2))} < 0$

Reorder the factors of $4 (3 (x - 2))$ .

$\frac{x (4 (x - 2))}{12 (x - 2)} - \frac{12}{12 (x - 2)} - \frac{(x + 1) (3 (x - 2))}{3 \cdot 4 (x - 2)} < 0$

Multiply $3$ by $4$ .

$\frac{x (4 (x - 2))}{12 (x - 2)} - \frac{12}{12 (x - 2)} - \frac{(x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

Combine the numerators over the common denominator.

$\frac{x (4 (x - 2)) - 12 - (x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

Simplify each term.

Rewrite using the commutative property of multiplication.

$\frac{4 x (x - 2) - 12 - (x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

Apply the distributive property.

$\frac{4 x \cdot x + 4 x \cdot - 2 - 12 - (x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

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

Move $x$ .

$\frac{4 (x \cdot x) + 4 x \cdot - 2 - 12 - (x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

Multiply $x$ by $x$ .

$\frac{4 x^{2} + 4 x \cdot - 2 - 12 - (x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

Multiply $- 2$ by $4$ .

$\frac{4 x^{2} - 8 x - 12 - (x + 1) (3 (x - 2))}{12 (x - 2)} < 0$

Apply the distributive property.

$\frac{4 x^{2} - 8 x - 12 + (- x - 1 \cdot 1) (3 (x - 2))}{12 (x - 2)} < 0$

Multiply $- 1$ by $1$ .

$\frac{4 x^{2} - 8 x - 12 + (- x - 1) (3 (x - 2))}{12 (x - 2)} < 0$

Apply the distributive property.

$\frac{4 x^{2} - 8 x - 12 + (- x - 1) (3 x + 3 \cdot - 2)}{12 (x - 2)} < 0$

Multiply $3$ by $- 2$ .

$\frac{4 x^{2} - 8 x - 12 + (- x - 1) (3 x - 6)}{12 (x - 2)} < 0$

Expand $(- x - 1) (3 x - 6)$ using the FOIL Method.

Apply the distributive property.

$\frac{4 x^{2} - 8 x - 12 - x (3 x - 6) - 1 (3 x - 6)}{12 (x - 2)} < 0$

Apply the distributive property.

$\frac{4 x^{2} - 8 x - 12 - x (3 x) - x \cdot - 6 - 1 (3 x - 6)}{12 (x - 2)} < 0$

Apply the distributive property.

$\frac{4 x^{2} - 8 x - 12 - x (3 x) - x \cdot - 6 - 1 (3 x) - 1 \cdot - 6}{12 (x - 2)} < 0$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$\frac{4 x^{2} - 8 x - 12 - 1 \cdot 3 x \cdot x - x \cdot - 6 - 1 (3 x) - 1 \cdot - 6}{12 (x - 2)} < 0$

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

Move $x$ .

$\frac{4 x^{2} - 8 x - 12 - 1 \cdot 3 (x \cdot x) - x \cdot - 6 - 1 (3 x) - 1 \cdot - 6}{12 (x - 2)} < 0$

Multiply $x$ by $x$ .

$\frac{4 x^{2} - 8 x - 12 - 1 \cdot 3 x^{2} - x \cdot - 6 - 1 (3 x) - 1 \cdot - 6}{12 (x - 2)} < 0$

Multiply $- 1$ by $3$ .

$\frac{4 x^{2} - 8 x - 12 - 3 x^{2} - x \cdot - 6 - 1 (3 x) - 1 \cdot - 6}{12 (x - 2)} < 0$

Multiply $- 6$ by $- 1$ .

$\frac{4 x^{2} - 8 x - 12 - 3 x^{2} + 6 x - 1 (3 x) - 1 \cdot - 6}{12 (x - 2)} < 0$

Multiply $3$ by $- 1$ .

$\frac{4 x^{2} - 8 x - 12 - 3 x^{2} + 6 x - 3 x - 1 \cdot - 6}{12 (x - 2)} < 0$

Multiply $- 1$ by $- 6$ .

$\frac{4 x^{2} - 8 x - 12 - 3 x^{2} + 6 x - 3 x + 6}{12 (x - 2)} < 0$

Subtract $3 x$ from $6 x$ .

$\frac{4 x^{2} - 8 x - 12 - 3 x^{2} + 3 x + 6}{12 (x - 2)} < 0$

Subtract $3 x^{2}$ from $4 x^{2}$ .

$\frac{x^{2} - 8 x - 12 + 3 x + 6}{12 (x - 2)} < 0$

Add $- 8 x$ and $3 x$ .

$\frac{x^{2} - 5 x - 12 + 6}{12 (x - 2)} < 0$

Add $- 12$ and $6$ .

$\frac{x^{2} - 5 x - 6}{12 (x - 2)} < 0$

Factor $x^{2} - 5 x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 5$ .

$- 6, 1$

Write the factored form using these integers.

$\frac{(x - 6) (x + 1)}{12 (x - 2)} < 0$

Step 3

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

$x - 6 = 0$

$x + 1 = 0$

$x - 2 = 0$

Step 4

Add $6$ to both sides of the equation.

$x = 6$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Add $2$ to both sides of the equation.

$x = 2$

Step 7

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

$x = 6$

$x = - 1$

$x = 2$

Step 8

Consolidate the solutions.

$x = 6, - 1, 2$

Step 9

Find the domain of $\frac{(x - 6) (x + 1)}{12 (x - 2)}$ .

Set the denominator in $\frac{(x - 6) (x + 1)}{12 (x - 2)}$ equal to $0$ to find where the expression is undefined.

$12 (x - 2) = 0$

Solve for $x$ .

Divide each term in $12 (x - 2) = 0$ by $12$ and simplify.

Divide each term in $12 (x - 2) = 0$ by $12$ .

$\frac{12 (x - 2)}{12} = \frac{0}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 (x - 2)}{12} = \frac{0}{12}$

Divide $x - 2$ by $1$ .

$x - 2 = \frac{0}{12}$

Simplify the right side.

Divide $0$ by $12$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

Step 10

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$2 < x < 6$

$x > 6$

Step 11

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 $x < - 1$ 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.

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

The left side $- 1.1\overline{6}$ is less than the right side $- 0.75$ , which means that the given statement is always true.

True

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.

$x = 0$

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

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

The left side $0.5$ is not less than the right side $0.25$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

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

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

The left side $0.8\overline{3}$ is less than the right side $1.25$ , which means that the given statement is always true.

True

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$\frac{8}{3} - \frac{1}{(8) - 2} < \frac{(8) + 1}{4}$

The left side $2.5$ is not less than the right side $2.25$ , which means that the given statement is false.

False

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

$x < - 1$ True

$- 1 < x < 2$ False

$2 < x < 6$ True

$x > 6$ False

$x < - 1$ True

$- 1 < x < 2$ False

$2 < x < 6$ True

$x > 6$ False

Step 12

The solution consists of all of the true intervals.

$x < - 1$ or $2 < x < 6$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or 2 < x < 6$

Interval Notation:

$(- \infty, - 1) \cup (2, 6)$

Step 14

Solve for x x/3-1/(x-2)<(x+1)/4

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