BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 1, Problem 3P
To determine

To solve: The equation |2x1||x+5|=3.

Expert Solution

Answer to Problem 3P

Solution:

The values of x=73 and 9.

Explanation of Solution

Given:

The equation is |2x1||x+5|=3.

Definition used:

An absolute function is defined as, |x|={x   if x0x if x<0.

From the definition, the given equation is expressed as follows:

|2x1|={2x1       if 2x10(2x1) if 2x1<0 and |x+5|={x+5       if x+50(x+5) if x+5<0.

Simplify further as, |2x1|={2x1   if x122x+1 if x<12 and |x+5|={x+5    if x5x5 if x<5.

Therefore, there exist three cases such as, x<5,5x<12 and x12.

Case 1: If x<5, solve the given equation for x as follows.

|2x1||x+5|=32x+1(x5)=32x+1+x+5=3x=3

The value of x=3. Note that x=3 is not lie on the any of the interval x<5,5x<12 and x12. So, this won’t be considered as a solution.

Case 2: If 5x<12, solve the given equation for x as follows.

|2x1||x+5|=32x+1(x+5)=32x+1x5=33x=7

The value of x=73.

Case 3: If x12, solve the given equation for x as follows.

|2x1||x+5|=3(2x1)(x+5)=32x1x5=3x=9

The value of x=9.

Combine the cases 1, 2, and 3, the values of x=73 and 9.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!