Graph Multi-Step Absolute Values Inequalities on a Number Line

Graph Multi-Step Absolute Value Inequalities on a Number Line - How it Works - Video

Example 1 Part 1

Example 1 Part 1:

When dealing with inequalities that have absolute value, we must isolate the absolute value or remove all operations that are outside the absolute value.

So with example we only have one operation that is outside the absolute value, which is multiplication. Now, we must divide since multiplication and division are inverses.

2 * | x - 4 | < 6

|x - 4 | < 3

Rewrite the inequality

Divide both sides by 2

Now since we have solved for the absolute value and it is on the left side of the inequality and we have less than or less than or equal to, we have an and statement.

Since we have an and statement, we have |x - 4 | < 3 and |x - 4 | > -3. We have flipped the second inequality; we flipped the sign and the number.

Remember absolute value is a distance so we went right to the positive number, 3, and now we need to go left to the negative number, -3.

| x - 4 | < 3

x - 4 < 3

x < 7

Rewrite the inequality

Drop absolute value signs

Add 4 to both sides

| x - 4 | > -3

x - 4 > -3

x > 1

Rewrite the inequality

Drop absolute value signs

Add 4 to both sides

Since we have an and statement, we need to combine the two inequalities together. We need to put the smaller number first then the bigger number so our compound inequality is 1 < x < 7.

Example 1 Part 2

Example 1 Part 2:

Now we graph our compound inequality. Since we have less than, our circles around the numbers are empty or open circles. Since we have an and statement, we are looking for the both inequalities to be true. So only the intersection of the two different inequalities, make the original inequality true.


Now, let's take a closer look at the graph. We have circled the number 4, because it is the average of the two numbers 1 and 7. If we count the number of spaces to the left and to the right, it is the same, 3 in this case. Do we see the number 4 somewhere? Do we the number 3 somewhere?

Example 1 Part 3

Example 1 Part 3:

The last part we are going to discuss is that our compound inequality is also our domain or our input values. We can also write using interval notation. Since we have less than, we put parenthesis around the numbers.

Example 2 Part 1

Example 2 Part 1:

When dealing with inequalities that have absolute value, we must isolate the absolute value or remove all operations that are outside the absolute value.

So with this example we have two operations that is outside the absolute value, which are multiplication and subtraction. Now, we must use addition then division to get the absolute value by itself.

3 * | x - 5 | - 2 < 7

3 * |x - 5 | < 9

| x - 5 | < 3

Rewrote the inequality

Add 2 to both sides

Divide both sides by 3

Now since we have solved for the absolute value and it is on the left side of the inequality and we have less than or less than or equal to, we have an and statement.

Since we have an and statement, we have |x - 5 | < 3 and |x - 5 | > -3. We have flipped the second inequality; we flipped the sign and the number.

Remember absolute value is a distance so we went right to the positive number, 3, and now we need to go left to the negative number, -3.

| x - 5 | < 3

x - 5 < 8

x < 8

Rewrite the inequality

Drop absolute value signs

Add 5 to both sides

| x - 4 | > -3

x - 5 > -3

x > 2

Rewrite the inequality

Drop absolute value signs

Add 5 to both sides

Since we have an and statement, we need to combine the two inequalities together. We need to put the smaller number first then the bigger number so our compound inequality is 2 < x < 8.

Example 2 Part 2

Example 2 Part 2:

Now we graph our compound inequality. Since we have less than, our circles around the numbers are empty or open circles. Since we have an and statement, we are looking for the both inequalities to be true. So only the intersection of the two different inequalities, make the original inequality true.


Now, let's take a closer look at the graph. We have circled the number 5, because it is the average of the two numbers 2 and 8. If we count the number of spaces to the left and to the right, it is the same, 3 in this case. Do we see the number 5 somewhere? Do we the number 3 somewhere?

Example 2 Part 3

Example 2 Part 3:

The last part we are going to discuss is that our compound inequality is also our domain or our input values. We can also write using interval notation. Since we have less than, we put parenthesis around the numbers.

Example 3 Part 1

Example 3 Part 1:

When dealing with inequalities that have absolute value, we must isolate the absolute value or remove all operations that are outside the absolute value.

So with this example we have two operations that is outside the absolute value, which are multiplication and subtraction. Now, we must use addition then division to get the absolute value by itself.

(1/4) * | x + 2 | - 1 ≥ 3

(1/4) * | x + 2 | 4

| x + 2 | 16

Rewrite the inequality

Add 1 to both sides

Multiply both sides by 4

Now since we have solved for the absolute value and it is on the left side of the inequality and we have greater than or greater than or equal to, we have an or statement.

Since we have an or statement, we have | x + 2 | 16 or | x + 2 | -16. We have flipped the second inequality; we flipped the sign and the number.

Remember absolute value is a distance so we went right to the positive number, 16, and now we need to go left to the negative number, -16.

| x + 2 | 16

x + 2 16

x 14

Rewrite the inequality

Drop absolute value signs

Subtract 2 to both sides

| x + 2 | -16

x + 2 -16

x -18

Rewrite the inequality

Drop absolute value signs

Subtract 2 to both sides

Since we have an or statement, we don't need to combine the two inequalities together. Although we don't need to combine them, we are going to rewrite it so the smaller number is first. So for our compound inequality, we have x -18 or 14 x. Notice we flipped the second inequality. We did that so the numbers are inside or closer together.

Example 3 Part 2

Example 3 Part 2:

Now we graph our compound inequality. Since we have less than or equal to, our circles around the numbers are filled or closed circles. Since we have an or statement, we are looking for any answer to be true. So we shade both sides since only one needs to be true at one time.


Now, let's take a closer look at the graph. We have circled the number 4, because it is the average of the two numbers 1 and 7. If we count the number of spaces to the left and to the right, it is the same, 3 in this case. Do we see the number 4 somewhere? Do we the number 3 somewhere?

Example 3 Part 3

Example 3 Part 3:

The last part we are going to discuss is that our compound inequality is also our domain or our input values. We can also write using interval notation. Since we have less than or equal to and an or statement, we put brackets (for the ... or equal to) around the numbers. And for the or statement we put the union symbol. Finally the since the arrows keep going on both sides we have positive infinity and negative infinity. Infinity always has parenthesis around when dealing with interval notation.

Example 4 Part 1

Example 4 Part 1:

When dealing with inequalities that have absolute value, we must isolate the absolute value or remove all operations that are outside the absolute value.

So with this example we have two operations that is outside the absolute value, which are multiplication, subtraction, and a hidden positive sign in front of the 2. Now, we must use subtraction then division to get the absolute value by itself.

2 - 5 * | x - 1 | > 12

- 5 * | x - 1 | > 10

| x - 1| < -2

Rewrite the inequality

Subtract 2 to both sides

Divide both sides by -5 and flip sign

Now since we have solved for the absolute value and it is on the left side of the inequality and we have less than or less than or equal to, we have an and statement.

Since we have an and statement, we have || x - 1| < -2 and | x - 1| > 2. We have flipped the second inequality; we flipped the sign and the number.

Remember absolute value is a distance so we went left to the negative number, -2 and now we need to go right to the positive number 2.


WAIT A MINUTE. Something strange is happening here. Absolute value can't be negative. Since only negative numbers are less than -2, there aren't any numbers that make this inequality true.

Example 4 Part 2

Example 4 Part 2:

Now we graph our compound inequality. Since we have less than, our circles around the numbers are empty or open circles. Since we have an and statement, we are looking for the intersection of the two different inequalities which is in the middle.


Now, let's take a closer look at the graph. We have circled the number 4, because it is the average of the two numbers 1 and 7. If we count the number of spaces to the left and to the right, it is the same, 3 in this case. Do we see the number 4 somewhere? Do we the number 3 somewhere?

Example 4 Part 3

Example 4 Part 3:

Since no input values work, we don't have a domain, and our answer is no solution.

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