Transcript C. - mckenziemath

Five-Minute Check (over Lesson 5–2)
CCSS
Then/Now
Example 1: Real-World Example: Solve a Multi-Step
Inequality
Example 2: Inequality Involving a Negative Coefficient
Example 3: Write and Solve an Inequality
Example 4: Distributive Property
Example 5: Empty Set and All Reals
Over Lesson 5–2
A. {a | a > 64}
B. {a | a < 64}
C. {a | a > 4}
D. {a | a < 4}
Over Lesson 5–2
A. {p | p > 28 }
B. {p | p < 28 }
C.
D. {p | p > –28 }
Over Lesson 5–2
Solve –9v ≥ –108.
A. {v | v ≥ 99}
B. {v | v ≤ 12}
C. {v | v ≥ 12}
D. {v | v ≥ –12}
Over Lesson 5–2
A. {c | c ≤ –5}
B. {c | c ≤ –2}
C.
D.
Over Lesson 5–2
Which inequality represents one half of Dan’s
savings is less than $60.00?
A.
B.
C.
D.
Over Lesson 5–2
Marta wants to purchase charms for her necklace.
Each charm costs $1.59. She wants to spend no
more than $20 for the charms. Which inequality
represents this situation? How many charms can
Marta purchase?
A. 1.59 – c > 20; 22
B. c + 1.59 < 20; 18
C. 1.59c ≥ 20; 12
D. 1.59c ≤ 20; 12
Content Standards
A.CED.1 Create equations and inequalities
in one variable and use them to solve
problems.
A.REI.3 Solve linear equations and
inequalities in one variable, including
equations with coefficients represented by
letters.
Mathematical Practices
7 Look for and make use of structure.
Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State
School Officers. All rights reserved.
You solved multi-step equations.
• Solve linear inequalities involving more than
one operation.
• Solve linear inequalities involving the
Distributive Property.
• Multi-step inequalities can be solved by
undoing the operations in the same way you
would solve a multi-step equation.
• Example: 11y – 13 = 42
+ 13 + 13
11y = 55
11
11
y=5
However, instead of = sign you will have an
inequality sign
• Write and solve an inequality to find the sales
Mrs. Jones needs if she earns a monthly
salary of $2000 plus a 10% commission on
her sales. Her goal is to make at least $4000
per month. What sales does she need to meet
the goal?
• Base salary + (commission x sales) ≥ 4000
2000 + 0.10x ≥ 4000
-2000
-2000
0.10x ≥ 2000
0.10 0.10
x ≥ 20000
Solve a Multi-Step Inequality
FAXES Adriana has a budget of $115 for faxes.
The fax service she uses charges $25 to activate
an account and $0.08 per page to send faxes. How
many pages can Adriana fax and stay within her
budget? Use the inequality 25 + 0.08p ≤ 115.
Original inequality
Subtract 25 from each side.
Divide each side by 0.08.
Simplify.
Answer: Adriana can send at most 1125 faxes.
• The Print Shop advertises a special to print
400 flyers for less than the competition. The
price includes a $3.50 set-up fee. If the
competition charges $35.50, what does the
Print Shop charge for each flyer?
• 400n + 3.50 < 35.50
- 3.50 - 3.50
400n < 32
400
400
n < $0.08
Rob has a budget of $425 for senior pictures. The
cost for a basic package and sitting fee is $200. He
wants to buy extra wallet-size pictures for his
friends that cost $4.50 each. How many wallet-size
pictures can he order and stay within his budget?
Use the inequality 200 + 4.5p ≤ 425.
A. 50 pictures
B. 55 pictures
C. 60 pictures
D. 70 pictures
• When multiplying or dividing by a negative the
direction of the inequality changes. This holds
true for multi-step inequalities.
• -11y – 13 > 42
+ 13 +13
-11y > 55
-11
-11
y < -5
The solution set is { y│y < -5}
Inequality Involving a Negative Coefficient
Solve 13 – 11d ≥ 79.
13 – 11d ≥ 79
13 – 11d – 13 ≥ 79 – 13
–11d ≥ 66
Original inequality
Subtract 13 from each side.
Simplify.
Divide each side by –11 and
change ≥ to ≤.
d ≤ –6
Simplify.
Answer: The solution set is {d | d ≤ –6} .
• 23 ≥ 10 – 2w
-10 -10
13 ≥ -2w
-2 -2
-13/2 ≤ w
The solution set is {w│w ≥ -13/2}
• 43 > -4y + 11
-11
- 11
32 > -4y
-4
-4
-8 < y
The solution set is {y│y > -8}
Solve –8y + 3 > –5.
A. {y | y < –1}
B. {y | y > 1}
C. {y | y > –1}
D. {y | y < 1}
• You can translate sentences
into multi-step inequalities and
then solve them using the
Properties of Inequalities
Write and Solve an Inequality
Define a variable, write an inequality, and solve the
problem below.
Four times a number plus twelve is less than the
number minus three.
Four
times a
number
plus
twelve
4n
+
12
is less than
<
a number
minus three.
n–3
Write and Solve an Inequality
4n + 12 < n – 3
Original inequality
4n + 12 – n < n – 3 – n Subtract n from each side.
3n + 12 < –3
3n + 12 – 12 < –3 – 12
3n < –15
Simplify.
Subtract 12 from each side.
Simplify.
Divide each side by 3.
n < –5
Simplify.
Answer: The solution set is {n | n < –5} .
• Five minus 6 times a number is more than
four times the number plus 45
• Five minus six times is more four times plus forty-five
a number
5 – 6n > 4n + 45
- 4n -4n
5 – 10n > 45
-5
-5
-10n > 40
-10
-10
n < -4
a number
• Two more than half a number is greater than
twenty-seven
• ½ n + 2 > 27
-2 -2
½ n > 25
(2)(1/2 n) > 25(2)
n > 50
{n│n > 50}
Write an inequality for the sentence below. Then
solve the inequality.
6 times a number is greater than 4 times the number
minus 2.
A. 6n > 4n – 2; {n | n > –1}
B. 6n < 4n – 2; {n | n < –1}
C. 6n > 4n + 2; {n | n > 1}
D. 6n > 2 – 4n;
• When solving inequalities that
contain grouping symbols, first use
the Distributive Property to remove
the grouping symbols. Next use
the order of operations to simplify
the resulting inequality.
Distributive Property
Solve 6c + 3(2 – c) ≥ –2c + 1.
6c + 3(2 – c) ≥ –2c + 1
Original inequality
6c + 6 – 3c ≥ –2c + 1
Distributive Property
3c + 6 ≥ –2c + 1
Combine like terms.
3c + 6 + 2c ≥ –2c + 1 + 2c Add 2c to each side.
5c + 6 ≥ 1
5c + 6 – 6 ≥ 1 – 6
5c ≥ –5
Simplify.
Subtract 6 from each side.
Simplify.
c ≥ –1
Divide each side by 5.
Answer: The solution set is {c | c ≥ –1}.
• 4(3t – 5) + 7 ≥ 8t + 3
12t – 20 + 7 ≥ 8t + 3
12t – 13 ≥ 8t + 3
-8t
-8t
4t – 13 ≥ 3
+13 +13
4t ≥ 16
4 4
t≥4
• Solve each inequality. Graph the solution on
a number line.
1) 6(5z – 3) ≤ 36z
30z – 18 ≤ 36z
-30z
-30z
-18 ≤ 6z
6
6
-3 ≤ z
The solution set is {z│z ≥ -3}
2) 2(h + 6) > -3(8 – h)
2h + 12 > -24 + 3h
-3h
- 3h
-h + 12 > -24
- 12 - 12
-h > -36
Remember the variable can not be negative
-1h > -36
-1
-1
h < 36
{h│h < 36}
Solve 3p – 2(p – 4) < p – (2 – 3p).
A.
p|p
B.
p|p
C.
D.
• If solving an inequality results in a statement
that is always true, the solution set is the set
of all real numbers. This solution set is
written as {x│x is a real number}
• If solving an inequality results in a statement
that is never true, the solution set is the
empty set, which is written as the symbol Ø.
• The empty set has no members.
Empty Set and All Reals
A. Solve –7(s + 4) + 11s ≥ 8s – 2(2s + 1).
–7(s + 4) + 11s ≥ 8s – 2(2s + 1) Original inequality
–7s – 28 + 11s ≥ 8s – 4s – 2
4s – 28 ≥ 4s – 2
4s – 28 – 4s ≥ 4s – 2 – 4s
– 28 ≥ – 2
Distributive Property
Combine like terms.
Subtract 4s from each
side.
Simplify.
Answer: Since the inequality results in a false
statement, the solution set is the empty set, Ø.
Empty Set and All Reals
B. Solve 2(4r + 3)  22 + 8(r – 2).
2(4r + 3) ≤ 22 + 8(r – 2) Original inequality
8r + 6 ≤ 22 + 8r – 16
Distributive Property
8r + 6 ≤ 6 + 8r
Simplify.
8r + 6 – 8r ≤ 6 + 8r – 8r
6≤ 6
Subtract 8r from each side.
Simplify.
Answer: All values of r make the inequality true.
All real numbers are the solution.
{r | r is a real number.}
• 9t – 5(t – 5) ≤ 4(t – 3)
9t – 5t + 25 ≤ 4t – 12
4t + 25 ≤ 4t – 12
-4t
-4t
25 ≤ -12
Since the inequality results in a false
statement, the solution set is the empty set Ø
• 3(4m + 6) ≤ 42 + 6(2m – 4)
12m + 18 ≤ 42 + 12m – 24
12m + 18 ≤ 12m + 18
-12m
-12m
18 ≤ 18
All values of m makes the inequality true. All real
numbers is the solution.
{m│m is a real numbers}
• Solve each inequality
1) 18 – 3(8c + 4) ≥ -6(4c – 1)
{c│c is a real number}
2) 46 ≤ 8m -4(2m + 5)
Ø
3) 3 – 8x ≥ 9 + 2(1 – 4x)
Ø
4) 3(2 – b) < 10 – 3(b – 6)
{b│b is a real number}
A. Solve 8a + 5 ≤ 6a + 3(a + 4) – (a + 7).
A. {a | a ≤ 3}
B. {a | a ≤ 0}
C. {a | a is a real number.}
D.
B. Solve 4r – 2(3 + r) < 7r – (8 + 5r).
A. {r | r > 0}
B. {r | r < –1}
C. {r | r is a real number.}
D.