Chapter 4: Q. 53 (page 242)

Solve each inequality algebraically.
$\frac{x + 1}{x - 3} \leq 2$

Short Answer

Expert verified

Solution set and the interval is:
${x | x < 3 o r x \geq 7} (- \infty, 3) \cup [7, \infty)$

Step by step solution

Step 1. Change the inequality to equal to zero.

Change the inequality equal to zero to make it easier and then get the value of x

Arrange the inequality first,

$\frac{x + 1}{x - 3} \leq 2 \frac{x + 1}{x - 3} - 2 \leq 0 \frac{x + 1}{x - 3} - 2 \cdot \frac{x - 3}{x - 3} \leq 0 \frac{x + 1}{x - 3} - \frac{2 x - 6}{x - 3} \leq 0 \frac{x + 1 - 2 x + 6}{x - 3} \leq 0 \frac{- x + 7}{x - 3} \leq 0$

Make it equal to zero,

$\frac{- x + 7}{x - 3} = 0 - x + 7 = 0; x - 3 = 0 x = 7; x = 3$

Step 2. Form the intervals

From the obtained values of x, we can form the interval,

So the interval we have is:

$(- \infty, 3) \cup (3, 7) \cup (7, \infty)$

Step 3. Form the table

Since, $f (x) < 0$ , so is negative.

check a number in each interval and evaluate the function to see whether it is satisfying the function.

Interval	Number chosen	Resultant
$(- \infty, 3)$	1	$- 3 \leq 0$
$(3, 7)$	4	$3 \leq 0$
$(7, \infty)$	8	$- \frac{1}{5} \leq 0$

Step 4. Solution set and interval

Since,

$- 3 \leq 0, - \frac{1}{5} \leq 0$ satisfy $f$ . Thus,

${x | x < 3 o r x \geq 7} (- \infty, 3) \cup [7, \infty)$

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Solve each inequality algebraically.
$\frac{x + 1}{x - 3} \leq 2$

Short Answer

Step by step solution

Step 1. Change the inequality to equal to zero.

Step 2. Form the intervals

Step 3. Form the table

Step 4. Solution set and interval

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Solve each inequality algebraically.x+1x-3≤2

Short Answer

Step by step solution

Step 1. Change the inequality to equal to zero.

Step 2. Form the intervals

Step 3. Form the table

Step 4. Solution set and interval

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Geometry

Logic and Functions

Calculus

Decision Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

Solve each inequality algebraically.
$\frac{x + 1}{x - 3} \leq 2$