Solutions to Problem Set 5

Math 101-06

9-13-2017

[Absolute value equations and inequalities]

1. Solve for x:

$$|x - 9| = 13.$$

$$\matrix{ & & |x - 9| = 13 & & \cr & \swarrow & & \searrow & \cr x - 9 = 13 & & & & x - 9 = -13 \cr x = 22 & & & & x = -4 \quad\halmos \cr}$$


2. Solve for x:

$$|x + 8| = -10.$$

An absolute value can't be negative. The equation has no solutions.


3. Solve for x:

$$|3 x - 5| = 10.$$

$$\matrix{ & & |3 x - 5| = 10 & & \cr & \swarrow & & \searrow & \cr 3 x - 5 = 10 & & & & 3 x - 5 = -10 \cr 3 x = 15 & & & & 3 x = -5 \cr \noalign{\vskip2pt} \cr x = 5 & & & & x = -\dfrac{5}{3} \quad\halmos \cr}$$


4. Solve $|x - 1| > 4$ for x. Write your answer using either interval notation or inequality notation.

Solve the corresponding equation:

$$\matrix{ & & |x - 1| = 4 & & \cr & \swarrow & & \searrow & \cr x - 1 = 4 & & & & x - 1 = -4 \cr x = 5 & & & & x = -3 \cr}$$

I put -3 and 5 on the number line. Since the absolute value expression is on the greater-than ("open") side of the "$>$ ", I shade the outside intervals:

$$\hbox{\epsfxsize=3in \epsffile{absolute-value-equations-and-inequalities-brackets1.eps}}$$

The solution is $x < -3$ or $x > 5$ .

In interval notation, this is $(-\infty, -3) \cup (5, \infty)$ .


5. Solve $|x + 2| < 10$ for x. Write your answer using either interval notation or inequality notation.

First, solve the helper equation $|x + 2| = 10$ .

$$\matrix{ & & |x + 2| = 10 & & \cr & & x + 2 = \pm 10 & & \cr & \swarrow & & \searrow & \cr x + 2 = 10 & & & & x + 2 = -10 \cr x = 8 & & & & x = -12 \cr}$$

I put -12 and 8 on the number line. Since the absolute value expression is on the less-than ("pointy") side of the "$<$ ", I shade the inside interval:

$$\hbox{\epsfxsize=3in \epsffile{absolute-value-equations-and-inequalities-brackets4.eps}}$$

The solution is $-12 < x < 8$ .

In interval notation, this is $(-12, 8)$ .


6. Solve $|x + 1| < -2$ for x. Write your answer using either interval notation or inequality notation.

An absolute value can't be negative. Hence, there are no solutions.


7. Solve $|17 x - 5| > -5$ for x. Write your answer using either interval notation or inequality notation.

An absolute value is always greater than or equal to 0. So $|17 x - 5|$ is always greater than a negative number like -5, no matter what x is. Hence, the solution is all real numbers (or $-\infty < x < \infty$ , or $(-\infty, \infty)$ ).


8. Solve $|2 x - 5| \le 7$ for x. Write your answer using either interval notation or inequality notation.

Solve the corresponding equation:

$$\matrix{ & & |2 x - 5| = 7 & & \cr & \swarrow & & \searrow & \cr 2 x - 5 = 7 & & & & 2 x - 5 = -7 \cr 2 x = 12 & & & & 2 x = -2 \cr x = 6 & & & & x = -1 \cr}$$

I put -1 and 6 on the number line. Since the absolute value expression is on the less-than ("pointy") side of the "$\le$ ", I shade the inside interval:

$$\hbox{\epsfxsize=3in \epsffile{absolute-value-equations-and-inequalities-brackets2.eps}}$$

The solution is $-1 \le x \le 6$ .

In interval notation, this is $[-1, 6]$ .


9. Solve $|10 - x| < 21$ for x. Write your answer using either interval notation or inequality notation.

Solve the corresponding equation:

$$\matrix{ & & |10 - x| = 21 & & \cr & \swarrow & & \searrow & \cr 10 - x = 21 & & & & 10 - x = -21 \cr -x = 11 & & & & -x = -31 \cr x = -11 & & & & x = 31 \cr}$$

I put -11 and 31 on the number line. Since the absolute value expression is on the less-than ("pointy") side of the "$<$ ", I shade the inside interval:

$$\hbox{\epsfxsize=3in \epsffile{absolute-value-equations-and-inequalities-brackets3.eps}}$$

The solution is $-11 < x < 31$ .

In interval notation, this is $(-11, 31)$ .


We are what we think. All that we are arises with our thoughts. With our thoughts we make the world. - The Buddha


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga