lesson 1.7, for use with pages 51-58 answer 1.3x + 15 = –42 2.5x – 8 ≤ 7 answer solve the...
TRANSCRIPT
Lesson 1.7, For use with pages 51-58
ANSWER
1. 3x + 15 = –42
2. 5x – 8 ≤ 7
ANSWER
Solve the equation or inequality.
–19
x ≤ 3
Lesson 1.7, For use with pages 51-58
3. 2x + 1 < –3 or 2x + 1 > 5
Solve the equation or inequality.
ANSWER x < –2 or x > 2
4. In the next 2 weeks you need to work at least 30 hours. If you can work h hours this week and then twice as many hours next week, how many hours must you work this week?
ANSWER at least 10 h
EXAMPLE 1 Solve a simple absolute value equation
Solve |x – 5| = 7. Graph the solution.
SOLUTION
| x – 5 | = 7
x – 5 = – 7 or x – 5 = 7
x = 5 – 7 or x = 5 + 7
x = –2 or x = 12
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
EXAMPLE 1
The solutions are –2 and 12. These are the values of x that are 7 units away from 5 on a number line. The graph is shown below.
ANSWER
Solve a simple absolute value equation
EXAMPLE 2 Solve an absolute value equation
| 5x – 10 | = 45
5x – 10 = 45 or 5x –10 = – 45
5x = 55 or 5x = – 35
x = 11 or x = – 7
Write original equation.
Expression can equal 45 or – 45 .
Add 10 to each side.
Divide each side by 5.
Solve |5x – 10 | = 45.
SOLUTION
EXAMPLE 2 Solve an absolute value equation
The solutions are 11 and –7. Check these in the original equation.
ANSWER
Check:
| 5x – 10 | = 45
| 5(11) – 10 | = 54?
|45| = 45?
45 = 45
| 5x – 10 | = 45
| 5(– 7 ) – 10 | = 54?
45 = 45
| – 45| = 45?
EXAMPLE 3
| 2x + 12 | = 4x
2x + 12 = 4x or 2x + 12 = – 4x
12 = 2x or 12 = – 6x
6 = x or –2 = x
Write original equation.
Expression can equal 4x or – 4 x
Add – 2x to each side.
Solve |2x + 12 | = 4x. Check for extraneous solutions.
SOLUTION
Solve for x.
Check for extraneous solutions
EXAMPLE 3
| 2x + 12 | = 4x
| 2(– 2) +12 | = 4(–2)?
|8| = – 8?
8 = –8
Check the apparent solutions to see if either is extraneous.
Check for extraneous solutions
| 2x + 12 | = 4x
| 2(6) +12 | = 4(6)?
|24| = 24?
24 = 24
The solution is 6. Reject – 2 because it is an extraneous solution.
ANSWER
CHECK
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
1. | x | = 5
SOLUTION
| x | = 5
| x | = – 5 or | x | = 5
x = –5 or x = 5
Write original equation.
Write equivalent equations.
Solve for x.
for Examples 1, 2 and 3
GUIDED PRACTICE for Examples 1, 2 and 3
The solutions are –5 and 5. These are the values of x that are 5 units away from 0 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1
0
1 2
3
4
5
6
7
– 5
– 6
– 7
5 5
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
2. |x – 3| = 10
SOLUTION
| x – 3 | = 10
x – 3 = – 10 or x – 3 = 10
x = 3 – 10 or x = 3 + 10
x = –7 or x = 13
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
for Examples 1, 2 and 3
GUIDED PRACTICE
The solutions are –7 and 13. These are the values of x that are 10 units away from 3 on a number line. The graph is shown below.
ANSWER
– 3
– 4
– 2
– 1
0
1
2
3
4
5
6
7
– 5
– 6
– 7
8
9
10
11
12
13
10 10
for Examples 1, 2 and 3
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
SOLUTION
| x + 2 | = 7
x + 2 = – 7 or x + 2 = 7
x = – 7 – 2 or x = 7 – 2
x = –9 or x = 5
Write original equation.
Write equivalent equations.
Solve for x.
Simplify.
3. |x + 2| = 7
for Examples 1, 2 and 3
GUIDED PRACTICE
The solutions are –9 and 5. These are the values of x that are 7 units away from – 2 on a number line.
ANSWER
for Examples 1, 2 and 3
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
4. |3x – 2| = 13
SOLUTION
3x – 2 = 13 or 3x – 2 = – 13
Write original equation.
Solve for x.
Simplify.
|3x – 2| = 13
Write equivalent equations.
x = or x = 5 3–3
2
for Examples 1, 2 and 3
x = –13 + 2
3or x =
13 + 23
GUIDED PRACTICE
The solutions are – and 5.
ANSWER
33
2
for Examples 1, 2 and 3
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
5. |2x + 5| = 3x
| 2x + 5 | = 3x
2x + 5 = – 3x or 2x + 5 = 3x
x = 1 or x = 5
Write original equation.
Write Equivalent equations.
Simplify
SOLUTION
for Examples 1, 2 and 3
2x + 3x = 5 or 2x – 3x = –5
GUIDED PRACTICE
The solution of is 5. Reject 1 because it is an extraneous solution.
ANSWER
for Examples 1, 2 and 3
| 2x + 12 | = 4x
| 2(1) +12 | = 4(1)?
|14| = 4 ? 14 = –8
Check the apparent solutions to see if either is extraneous.
| 2x + 5 | = 3x
| 2(5) +5 | = 3(5) ?
|15| = 15?
15 = 15
CHECK
GUIDED PRACTICE
Solve the equation. Check for extraneous solutions.
6. |4x – 1| = 2x + 9
SOLUTION
4x – 1 = – (2x + 9) or 4x – 1 = 2x + 9
Write original equation.
Solve For xx = or x = 5 31
1
|4x – 1| = 2x + 9
Write equivalent equations.
for Examples 1, 2 and 3
4x + 2x = – 9 + 1 or 4x – 2x = 9 + 1 Rewrite equation.
GUIDED PRACTICE
ANSWER
The solutions are – and 5. 3
11
for Examples 1, 2 and 3
| 4x – 1 | = 2x + 9
| 4(5) – 1 | = 2(5) + 9?
|19| = 19 ?
19 = 19
Check the apparent solutions to see if either is extraneous.
| 4x – 1 | = 2x + 9
|4( -1 ) – 1 | = 2 ( - 1 )+ 931
3
1?
CHECK
| | = ? 3
– 19
319
= 3
– 19
319
EXAMPLE 4 Solve an inequality of the form |ax + b| > c
Solve |4x + 5| > 13. Then graph the solution.
SOLUTION
First Inequality Second Inequality
4x + 5 < – 13 4x + 5 > 13
4x < – 18 4x > 8
x < – 92
x > 2
Write inequalities.
Subtract 5 from each side.
Divide each side by 4.
The absolute value inequality is equivalent to
4x +5 < –13 or 4x + 5 > 13.
EXAMPLE 4
ANSWER
Solve an inequality of the form |ax + b| > c
The solutions are all real numbers less than or greater than 2. The graph is shown below.– 9
2
EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c
A professional baseball should weigh 5.125 ounces, with a tolerance of 0.125 ounce. Write and solve an absolute value inequality that describes the acceptable weights for a baseball.
Baseball
SOLUTION
Write a verbal model. Then write an inequality.STEP 1
EXAMPLE 5 Solve an inequality of the form |ax + b| ≤ c
STEP 2 Solve the inequality.
Write inequality.
Write equivalent compound inequality.
Add 5.125 to each expression.
|w – 5.125| ≤ 0.125
– 0.125 ≤ w – 5.125 ≤ 0.125
5 ≤ w ≤ 5.25
So, a baseball should weigh between 5 ounces and 5.25 ounces, inclusive. The graph is shown below.
ANSWER
EXAMPLE 6
The thickness of the mats used in the rings, parallel bars, and vault events must be between 7.5 inches and 8.25 inches, inclusive. Write an absolute value inequality describing the acceptable mat thicknesses.
Gymnastics
SOLUTION
STEP 1 Calculate the mean of the extreme mat thicknesses.
Write a range as an absolute value inequality
EXAMPLE 6
Mean of extremes = = 7.875 7.5 + 8.25
2
Find the tolerance by subtracting the mean from the upper extreme.
STEP 2
Tolerance = 8.25 – 7.875
Write a range as an absolute value inequality
= 0.375
EXAMPLE 6
STEP 3 Write a verbal model. Then write an inequality.
A mat is acceptable if its thickness t satisfies |t – 7.875| ≤ 0.375.
ANSWER
Write a range as an absolute value inequality
GUIDED PRACTICE for Examples 5 and 6
Solve the inequality. Then graph the solution.
10. |x + 2| < 6
The absolute value inequality is equivalent to x + 2 < 6 or x + 2 > – 6
First Inequality Second Inequality
x + 2 < 6 x + 2 > – 6
x < 4 x > – 8
Write inequalities.
Subtract 2 from each side.
SOLUTION
GUIDED PRACTICE for Examples 5 and 6
ANSWER
The solutions are all real numbers less than – 8 or greater than 4. The graph is shown below.
GUIDED PRACTICE for Examples 5 and 6
Solve the inequality. Then graph the solution.
11. |2x + 1| ≤ 9
The absolute value inequality is equivalent to 2x + 1 < 9 or 2x + 1 > – 9
First Inequality Second Inequality
2x < 8 2x > – 10
Write inequalities.
Subtract 1 from each side.
SOLUTION
2x + 1 > – 92x + 1 < 9
x < 4 x > – 5Divide each side 2
GUIDED PRACTICE for Examples 5 and 6
ANSWER
The solutions are all real numbers less than – 5 or greater than 4. The graph is shown below.
GUIDED PRACTICE for Examples 5 and 6
12. |7 – x| ≤ 4
Solve the inequality. Then graph the solution.
The absolute value inequality is equivalent to 7 – x < 4 or 7 – x > – 4
First Inequality Second Inequality
– x < – 3 – x > – 11
Write inequalities.
Subtract 7 from each side.
SOLUTION
x < 3 x > 11Divide each side (– ) sign
7 – x > – 47 – x < 4
GUIDED PRACTICE for Examples 5 and 6
ANSWER
The solutions are all real numbers less than 3 or greater than 11. The graph is shown below.
GUIDED PRACTICE for Examples 5 and 6
13. Gymnastics: For Example 6, write an absolute value inequality describing the unacceptable mat thicknesses.
SOLUTION
STEP 1 Calculate the mean of the extreme mat thicknesses.
Mean of extremes = = 7.875 7.5+ 8.25
2
Find the tolerance by subtracting the mean from the upper extreme.
STEP 2
Tolerance = 8.25 – 7.875 = 0.375
GUIDED PRACTICE for Examples 5 and 6
STEP 3
A mat is unacceptable if its thickness t satisfies |t – 7.875| > 0.375.
ANSWER
GUIDED PRACTICE for Example 4
Solve the inequality. Then graph the solution.
7. |x + 4| ≥ 6
x < – 10 or x > 2 The graph is shown below.
ANSWER
GUIDED PRACTICE for Example 4
Solve the inequality. Then graph the solution.
8. |2x –7|>1
ANSWER
x < 3 or x > 4 The graph is shown below.
GUIDED PRACTICE for Example 4
Solve the inequality. Then graph the solution.
9. |3x + 5| ≥ 10
ANSWER
x < – 5 or x > 123
The graph is shown below.