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Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. 1. Review of Solving Absolute Value Equations • Example: | x – 2 | = 10 Remember to drop the absolute value sign and set the left side of the equation equal to both 10 and -10 Then, solve for x in both equations x – 2 = 10 x – 2 = - 10 | x – 2 | = 10 OR +2 = +2 x = 12 +2 = +2 x = -8 OR

Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

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Page 1: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

1. 1. Review of Solving Absolute Value Equations

• Example: • | x – 2 | = 10

• Remember to drop the absolute value sign and set the left side of the equation equal to both 10 and -10

• Then, solve for x in both equations

x – 2 = 10 x – 2 = -10

| x – 2 | = 10

OR

+2 = +2

x = 12

+2 = +2

x = -8OR

Page 2: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

2. 2. Absolute Value Equations

WITH TERMS OUTSIDE ABSOLUTE VALUE SIGNS

• Sometimes there are terms outside of the absolute value sign

• First, isolate the terms in the absolute value sign

| x – 3 | + 4 = 8

x – 3 = 4 x – 3 = -4

| x – 3 | = 4

• Second, remove absolute value signs and solve two equations

OR

-4 = -4

Page 3: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

3. 3. Examples

x + 3 = 15 x + 3 = -15

Ex 2: | x + 3| – 5 = 10

OR

-3 = -3

x = 12

-3 = -3

x = -18OR

| 12 + 3 | – 5 = 10

CHECK:

| x + 3 | = 15

+5 = +5

| -18 + 3 | – 5 = 10

| 15 | – 5 = 10 | -15 | – 5 = 1015 – 5 = 10

Page 4: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

3. 3. Examples

x + 5 = 18 x + 5 = -18

Ex 3: | x + 5| + 7 = 25

OR

-5 = -5

x = 13

-5 = -5

x = -23OR

| 13 + 5 | + 7 = 25

CHECK:

| x + 5 | = 18

-7 = -7

| -23 + 5 | + 7 = 25

| 18 | + 7 = 25 | -18 | + 7 = 2518 + 7 = 25

Page 5: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

4. 4. Absolute Value Word Problems

The temperature at 6am was 37ºF. The temperature at 2pm was 65ºF. The temperature at 10pm was 33ºF. Using absolute value, determine if the temperature change in the morning was more than the change at night.

What are we trying to find?

What do we know?

If the temperature change in morning was more than the change at night

• Temp in ºF• 6am: 37, 2pm: 65,

10pm: 33• Use Absolute value• Chg in morn and night

IT SAYS I SAY• Morning is chg from 6am to

2pm• Night is chg from 2pm to

10pm• Abs must be positive

Page 6: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

4. 4. Absolute Value Word Problems

The temperature at 6am was 37ºF. The temperature at 2pm was 65ºF. The temperature at 10pm was 33ºF. Using absolute value, determine if the temperature change in the morning was more than the change at night.

How do we solve it? Did we answer the original question?

• Change in Morning:• | 65 – 37 | = |28| = 28ºF

• Change at Night:• | 33 – 65 | = |-32| = 32ºF

Chg at Morn < Chg at Night28ºF < 32ºF

• YES!

• The answer is no. The change at night was more than the change during the morning by 4ºF.

Page 7: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

4. 4. Absolute Value Word Problems

Uplift is on Wilson Ave, which is 4600 North (in grid coordinates). Brianna lives at 2000 North and Chris lives at 7000 North. Assume everyone lives on Sheridan/Sheffield. Using absolute value, determine who lives further from Uplift.

What are we trying to find?

What do we know?

Who lives further from Uplift (in grid coordinates)

• Uplift 4600 North• Brianna 2000 North• Chris 7000 North• Use Abs Value

IT SAYS I SAY

• Chicago map, in line north to south

• Grid coordinates measure blocks in Chicago

• Abs value is always positive

7000N

2000N

4600N

Page 8: Solving Absolute Value Equations and Word Problems Name May 12, 2011 1. Review of Solving Absolute Value Equations Example: | x – 2 | = 10 Remember to

Solving Absolute Value Equations andWord Problems

Name

May 12, 2011

4. 4. Absolute Value Word Problems

How do we solve it? Did we answer the original question?

• Brianna Distance from Uplift:• |2000 – 4600| = |-2600| = 2600

• Chris Distance from Uplift• |7000 - 4600| = |2400| = 2400

Brianna Dist. > Chris Dist.2600 > 2400

• YES!

• Brianna lives further from Uplift than Chris by 200

Uplift is on Wilson Ave, which is 4600 North (in grid coordinates). Brianna lives at 2000 North and Chris lives at 7000 North. Assume everyone lives on Sheridan/Sheffield. Using absolute value, determine who lives further from Uplift.

7000N

2000N

4600N

2400

2600