Rational Inequalities - College Algebra

Card 1 of 8

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 87 left

Question

Solve

$$\frac{x+2}{x-1}$\geq 0$

Tap to reveal answer

Answer

In order to solve this, we need to figure out what values of x makes the numerator and denominator equal zero.

These values are , and .

Then test a few numbers greater or lesser than the above values to figure out what range of values make the inequality true.

$-\infty< x \leq -2$

$1<x<\infty$

← Didn't Know|Knew It →

Question

Give the solution set of the equation

Tap to reveal answer

Answer

The boundary points of a rational inequality are the zeroes of the numerator and the denominator.

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are $\left \{ 0,1,7\right \}$ , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

$(-\infty, 0)$ : Set

$$\frac{4 $(-1)}{(-1)^{2}$$-8(-1) + 7} \le 0$

$$\frac{-4}{16}$ \le 0$

$-$\frac{1}{4}$ \le 0$

True; include $(-\infty, 0)$ .

: Set $x= $\frac{1}{2}$$

$$\frac{4\cdot \frac{1}{2}$}{ \left($\frac{1}{2}$ \right $)^{2}$-8 \cdot $\frac{1}{2}$ + 7} \le 0$

$$\frac{2}{ \frac{1}${4} -4 + 7} \le 0$

$$\frac{2}{ \frac{13}${4} } \le 0$

$$\frac{8}{ 13}$ \le 0$

False; exclude .

: Set :

$$\frac{8}{4-16 + 7}$ \le 0$

$$\frac{8}{-5}$ \le 0$

$-$\frac{8}{5}$ \le 0$

True; include

$(7, \infty)$ : Set :

$$\frac{4 \cdot 8 $}{8^{2}$$-8 \cdot 8 + 7} \le 0$

$$\frac{32}{64-64 + 7}$ \le 0$

$$\frac{32}{ 7}$ \le 0$

False; exclude $(7, \infty)$ .

Since the inequality symbol is the "is greater than or equal to" symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

$(-\infty, 0] \cup (1,7)$ .

← Didn't Know|Knew It →

Question

Solve

$$\frac{x+2}{x-1}$\geq 0$

Tap to reveal answer

Answer

In order to solve this, we need to figure out what values of x makes the numerator and denominator equal zero.

These values are , and .

Then test a few numbers greater or lesser than the above values to figure out what range of values make the inequality true.

$-\infty< x \leq -2$

$1<x<\infty$

← Didn't Know|Knew It →

Question

Give the solution set of the equation

Tap to reveal answer

Answer

The boundary points of a rational inequality are the zeroes of the numerator and the denominator.

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are $\left \{ 0,1,7\right \}$ , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

$(-\infty, 0)$ : Set

$$\frac{4 $(-1)}{(-1)^{2}$$-8(-1) + 7} \le 0$

$$\frac{-4}{16}$ \le 0$

$-$\frac{1}{4}$ \le 0$

True; include $(-\infty, 0)$ .

: Set $x= $\frac{1}{2}$$

$$\frac{4\cdot \frac{1}{2}$}{ \left($\frac{1}{2}$ \right $)^{2}$-8 \cdot $\frac{1}{2}$ + 7} \le 0$

$$\frac{2}{ \frac{1}${4} -4 + 7} \le 0$

$$\frac{2}{ \frac{13}${4} } \le 0$

$$\frac{8}{ 13}$ \le 0$

False; exclude .

: Set :

$$\frac{8}{4-16 + 7}$ \le 0$

$$\frac{8}{-5}$ \le 0$

$-$\frac{8}{5}$ \le 0$

True; include

$(7, \infty)$ : Set :

$$\frac{4 \cdot 8 $}{8^{2}$$-8 \cdot 8 + 7} \le 0$

$$\frac{32}{64-64 + 7}$ \le 0$

$$\frac{32}{ 7}$ \le 0$

False; exclude $(7, \infty)$ .

Since the inequality symbol is the "is greater than or equal to" symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

$(-\infty, 0] \cup (1,7)$ .

← Didn't Know|Knew It →

Question

Solve

$$\frac{x+2}{x-1}$\geq 0$

Tap to reveal answer

Answer

In order to solve this, we need to figure out what values of x makes the numerator and denominator equal zero.

These values are , and .

Then test a few numbers greater or lesser than the above values to figure out what range of values make the inequality true.

$-\infty< x \leq -2$

$1<x<\infty$

← Didn't Know|Knew It →

Question

Give the solution set of the equation

Tap to reveal answer

Answer

The boundary points of a rational inequality are the zeroes of the numerator and the denominator.

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are $\left \{ 0,1,7\right \}$ , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

$(-\infty, 0)$ : Set

$$\frac{4 $(-1)}{(-1)^{2}$$-8(-1) + 7} \le 0$

$$\frac{-4}{16}$ \le 0$

$-$\frac{1}{4}$ \le 0$

True; include $(-\infty, 0)$ .

: Set $x= $\frac{1}{2}$$

$$\frac{4\cdot \frac{1}{2}$}{ \left($\frac{1}{2}$ \right $)^{2}$-8 \cdot $\frac{1}{2}$ + 7} \le 0$

$$\frac{2}{ \frac{1}${4} -4 + 7} \le 0$

$$\frac{2}{ \frac{13}${4} } \le 0$

$$\frac{8}{ 13}$ \le 0$

False; exclude .

: Set :

$$\frac{8}{4-16 + 7}$ \le 0$

$$\frac{8}{-5}$ \le 0$

$-$\frac{8}{5}$ \le 0$

True; include

$(7, \infty)$ : Set :

$$\frac{4 \cdot 8 $}{8^{2}$$-8 \cdot 8 + 7} \le 0$

$$\frac{32}{64-64 + 7}$ \le 0$

$$\frac{32}{ 7}$ \le 0$

False; exclude $(7, \infty)$ .

Since the inequality symbol is the "is greater than or equal to" symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

$(-\infty, 0] \cup (1,7)$ .

← Didn't Know|Knew It →

Question

Solve

$$\frac{x+2}{x-1}$\geq 0$

Tap to reveal answer

Answer

In order to solve this, we need to figure out what values of x makes the numerator and denominator equal zero.

These values are , and .

Then test a few numbers greater or lesser than the above values to figure out what range of values make the inequality true.

$-\infty< x \leq -2$

$1<x<\infty$

← Didn't Know|Knew It →

Question

Give the solution set of the equation

Tap to reveal answer

Answer

The boundary points of a rational inequality are the zeroes of the numerator and the denominator.

First, set the numerator equal to 0 and solve for :

Now set the denominator equal to 0 and solve for :

Factor the trinomial using the reverse-FOIL method. Look for two integers whose sum is and whose product is 7; these are , and , so the equation can be rewritten as

Set each factor to zero and solve for :

Therefore, the boundary points are $\left \{ 0,1,7\right \}$ , which divide the real numbers into four intervals. Choose any value from each interval as a test point, setting to that value and determining whether the inequality is true.

The four intervals are listed below, along with their arbitrary test points.

$(-\infty, 0)$ : Set

$$\frac{4 $(-1)}{(-1)^{2}$$-8(-1) + 7} \le 0$

$$\frac{-4}{16}$ \le 0$

$-$\frac{1}{4}$ \le 0$

True; include $(-\infty, 0)$ .

: Set $x= $\frac{1}{2}$$

$$\frac{4\cdot \frac{1}{2}$}{ \left($\frac{1}{2}$ \right $)^{2}$-8 \cdot $\frac{1}{2}$ + 7} \le 0$

$$\frac{2}{ \frac{1}${4} -4 + 7} \le 0$

$$\frac{2}{ \frac{13}${4} } \le 0$

$$\frac{8}{ 13}$ \le 0$

False; exclude .

: Set :

$$\frac{8}{4-16 + 7}$ \le 0$

$$\frac{8}{-5}$ \le 0$

$-$\frac{8}{5}$ \le 0$

True; include

$(7, \infty)$ : Set :

$$\frac{4 \cdot 8 $}{8^{2}$$-8 \cdot 8 + 7} \le 0$

$$\frac{32}{64-64 + 7}$ \le 0$

$$\frac{32}{ 7}$ \le 0$

False; exclude $(7, \infty)$ .

Since the inequality symbol is the "is greater than or equal to" symbol, include the zero, 0, of the numerator - but not the other two boundaries, the zeroes of the denominator. This makes the solution set

$(-\infty, 0] \cup (1,7)$ .

← Didn't Know|Knew It →

0

Knew It