Convert to Interval Notation (7x+10)/(x-2)<=x-5

$\frac{7 x + 10}{x - 2} \leq x - 5$

Step 1

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the inequality.

$\frac{7 x + 10}{x - 2} - x \leq - 5$

Add $5$ to both sides of the inequality.

$\frac{7 x + 10}{x - 2} - x + 5 \leq 0$

Step 2

Simplify $\frac{7 x + 10}{x - 2} - x + 5$ .

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

$\frac{7 x + 10}{x - 2} - x \cdot \frac{x - 2}{x - 2} + 5 \leq 0$

Combine $- x$ and $\frac{x - 2}{x - 2}$ .

$\frac{7 x + 10}{x - 2} + \frac{- x (x - 2)}{x - 2} + 5 \leq 0$

Combine the numerators over the common denominator.

$\frac{7 x + 10 - x (x - 2)}{x - 2} + 5 \leq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{7 x + 10 - x \cdot x - x \cdot - 2}{x - 2} + 5 \leq 0$

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

Move $x$ .

$\frac{7 x + 10 - (x \cdot x) - x \cdot - 2}{x - 2} + 5 \leq 0$

Multiply $x$ by $x$ .

$\frac{7 x + 10 - x^{2} - x \cdot - 2}{x - 2} + 5 \leq 0$

Multiply $- 2$ by $- 1$ .

$\frac{7 x + 10 - x^{2} + 2 x}{x - 2} + 5 \leq 0$

Add $7 x$ and $2 x$ .

$\frac{9 x + 10 - x^{2}}{x - 2} + 5 \leq 0$

Reorder terms.

$\frac{- x^{2} + 9 x + 10}{x - 2} + 5 \leq 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 10 = - 10$ and whose sum is $b = 9$ .

Factor $9$ out of $9 x$ .

$\frac{- x^{2} + 9 (x) + 10}{x - 2} + 5 \leq 0$

Rewrite $9$ as $- 1$ plus $10$

$\frac{- x^{2} + (- 1 + 10) x + 10}{x - 2} + 5 \leq 0$

Apply the distributive property.

$\frac{- x^{2} - 1 x + 10 x + 10}{x - 2} + 5 \leq 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(- x^{2} - 1 x) + 10 x + 10}{x - 2} + 5 \leq 0$

Factor out the greatest common factor (GCF) from each group.

$\frac{x (- x - 1) - 10 (- x - 1)}{x - 2} + 5 \leq 0$

Factor the polynomial by factoring out the greatest common factor, $- x - 1$ .

$\frac{(- x - 1) (x - 10)}{x - 2} + 5 \leq 0$

To write $5$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{(- x - 1) (x - 10)}{x - 2} + \frac{5 (x - 2)}{x - 2} \leq 0$

Combine the numerators over the common denominator.

$\frac{(- x - 1) (x - 10) + 5 (x - 2)}{x - 2} \leq 0$

Simplify the numerator.

Expand $(- x - 1) (x - 10)$ using the FOIL Method.

Apply the distributive property.

$\frac{- x (x - 10) - 1 (x - 10) + 5 (x - 2)}{x - 2} \leq 0$

Apply the distributive property.

$\frac{- x \cdot x - x \cdot - 10 - 1 (x - 10) + 5 (x - 2)}{x - 2} \leq 0$

Apply the distributive property.

$\frac{- x \cdot x - x \cdot - 10 - 1 x - 1 \cdot - 10 + 5 (x - 2)}{x - 2} \leq 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$\frac{- (x \cdot x) - x \cdot - 10 - 1 x - 1 \cdot - 10 + 5 (x - 2)}{x - 2} \leq 0$

Multiply $x$ by $x$ .

$\frac{- x^{2} - x \cdot - 10 - 1 x - 1 \cdot - 10 + 5 (x - 2)}{x - 2} \leq 0$

Multiply $- 10$ by $- 1$ .

$\frac{- x^{2} + 10 x - 1 x - 1 \cdot - 10 + 5 (x - 2)}{x - 2} \leq 0$

Rewrite $- 1 x$ as $- x$ .

$\frac{- x^{2} + 10 x - x - 1 \cdot - 10 + 5 (x - 2)}{x - 2} \leq 0$

Multiply $- 1$ by $- 10$ .

$\frac{- x^{2} + 10 x - x + 10 + 5 (x - 2)}{x - 2} \leq 0$

Subtract $x$ from $10 x$ .

$\frac{- x^{2} + 9 x + 10 + 5 (x - 2)}{x - 2} \leq 0$

Apply the distributive property.

$\frac{- x^{2} + 9 x + 10 + 5 x + 5 \cdot - 2}{x - 2} \leq 0$

Multiply $5$ by $- 2$ .

$\frac{- x^{2} + 9 x + 10 + 5 x - 10}{x - 2} \leq 0$

Add $9 x$ and $5 x$ .

$\frac{- x^{2} + 14 x + 10 - 10}{x - 2} \leq 0$

Subtract $10$ from $10$ .

$\frac{- x^{2} + 14 x + 0}{x - 2} \leq 0$

Add $- x^{2} + 14 x$ and $0$ .

$\frac{- x^{2} + 14 x}{x - 2} \leq 0$

Factor $x$ out of $- x^{2} + 14 x$ .

Factor $x$ out of $- x^{2}$ .

$\frac{x (- x) + 14 x}{x - 2} \leq 0$

Factor $x$ out of $14 x$ .

$\frac{x (- x) + x \cdot 14}{x - 2} \leq 0$

Factor $x$ out of $x (- x) + x \cdot 14$ .

$\frac{x (- x + 14)}{x - 2} \leq 0$

Factor $- 1$ out of $- x$ .

$\frac{x (- (x) + 14)}{x - 2} \leq 0$

Rewrite $14$ as $- 1 (- 14)$ .

$\frac{x (- (x) - 1 (- 14))}{x - 2} \leq 0$

Factor $- 1$ out of $- (x) - 1 (- 14)$ .

$\frac{x (- (x - 14))}{x - 2} \leq 0$

Rewrite $- (x - 14)$ as $- 1 (x - 14)$ .

$\frac{x (- 1 (x - 14))}{x - 2} \leq 0$

Move the negative in front of the fraction.

$- \frac{x (x - 14)}{x - 2} \leq 0$

Step 3

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

$- x = 0$

$x - 14 = 0$

$x - 2 = 0$

Step 4

Divide each term in $- x = 0$ by $- 1$ and simplify.

Divide each term in $- x = 0$ by $- 1$ .

$\frac{- x}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{0}{- 1}$

Divide $x$ by $1$ .

$x = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$x = 0$

Step 5

Add $14$ to both sides of the equation.

$x = 14$

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.

$x = 0$

$x = 14$

$x = 2$

Step 8

Consolidate the solutions.

$x = 0, 14, 2$

Step 9

Find the domain of $- \frac{x (x - 14)}{x - 2}$ .

Set the denominator in $\frac{x (x - 14)}{x - 2}$ equal to $0$ to find where the expression is undefined.

$x - 2 = 0$

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.

$x < 0$

$0 < x < 2$

$2 < x < 14$

$x > 14$

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 $x < 0$ to see if it makes the inequality true.

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

$x = - 2$

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

$\frac{7 (- 2) + 10}{(- 2) - 2} \leq (- 2) - 5$

The left side $1$ is greater than the right side $- 7$ , which means that the given statement is false.

False

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

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

$x = 1$

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

$\frac{7 (1) + 10}{(1) - 2} \leq (1) - 5$

The left side $- 17$ is less than the right side $- 4$ , which means that the given statement is always true.

True

Test a value on the interval $2 < x < 14$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 14$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{7 (4) + 10}{(4) - 2} \leq (4) - 5$

The left side $19$ is greater than the right side $- 1$ , which means that the given statement is false.

False

Test a value on the interval $x > 14$ to see if it makes the inequality true.

Choose a value on the interval $x > 14$ and see if this value makes the original inequality true.

$x = 16$

Replace $x$ with $16$ in the original inequality.

$\frac{7 (16) + 10}{(16) - 2} \leq (16) - 5$

The left side $8.\overline{714285}$ is less than the right side $11$ , which means that the given statement is always true.

True

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

$x < 0$ False

$0 < x < 2$ True

$2 < x < 14$ False

$x > 14$ True

$x < 0$ False

$0 < x < 2$ True

$2 < x < 14$ False

$x > 14$ True

Step 12

The solution consists of all of the true intervals.

$0 \leq x < 2$ or $x \geq 14$

Step 13

Convert the inequality to interval notation.

$[0, 2) \cup [14, \infty)$

Step 14

Convert to Interval Notation (7x+10)/(x-2)<=x-5

Related Questions

Support LCMGCF.com by disabling your adblocker.