Transcript 3-5 day 1

Inequalities with
3-5 Solving
Variables on Both Sides
Warm Up
Solve each equation.
1. 2x = 7x + 15 x = –3
2. 3y – 21 = 4 – 2y
y=5
3. 2(3z + 1) = –2(z + 3) z = –1
4. 3(p – 1) = 3p + 2
no solution
5. Solve and graph 5(2 – b) > 52. b < –3
–6
–5
Holt McDougal Algebra 1
–4
–3
–2
–1
0
Inequalities with
3-5 Solving
Variables on Both Sides
Objective: to solve multistep inequalities
DO NOW:
Complete “Error Analysis” worksheet.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Review Do Now
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Solving Inequalities with Variables
on Both Sides –In your notes!
Some inequalities have variable terms on both
sides of the inequality symbol. You can solve
these inequalities just like you solved
equations with variables on both sides.
Use the properties of inequality to “collect” all
the variable terms on one side and all the
constant terms on the other side.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 1:
Solve the inequality and graph the solutions.
4m – 3 < 2m + 6
–2m
– 2m
2m – 3 <
+6
+3
+3
2m
<
9
To collect the variable terms on one
side, subtract 2m from both sides.
Since 3 is subtracted from 2m, add
3 to both sides to undo the
subtraction
Since m is multiplied by 2, divide
both sides by 2 to undo the
multiplication.
4
Holt McDougal Algebra 1
5
6
Inequalities with
3-5 Solving
Variables on Both Sides
You may need to simplify one or both sides
of an inequality before solving it. Look for like
terms to combine and places to use the
Distributive Property.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 2:
Solve the inequality and graph the solutions.
2(k – 3) > 6 + 3k – 3
2(k – 3) > 3 + 3k
2k + 2(–3) > 3 + 3k
2k – 6 > 3 + 3k
–2k
– 2k
Distribute 2 on the left side of
the inequality.
To collect the variable terms,
subtract 2k from both
sides.
–6 > 3 + k
–3
–3
–9 > k
So, k<-9
Holt McDougal Algebra 1
–12
–9
–6
–3
0
3
Inequalities with
3-5 Solving
Variables on Both Sides
Example 3
Solve the inequality and graph the solutions.
5(2 – r) ≥ 3(r – 2)
Distribute 5 on the left side of the
5(2 – r) ≥ 3(r – 2)
inequality and distribute 3 on
5(2) – 5(r) ≥ 3(r) + 3(–2) the right side of the inequality.
10 – 5r ≥ 3r – 6
+ 5r +5r
10 ≥ 8r - 6
+6
+6
16 ≥ 8r
2≥r
Holt McDougal Algebra 1
So,
r≤2
–6
–4
–2
0
2
4
Inequalities with
3-5 Solving
Variables on Both Sides
You try!
Solve each inequality and graph the solutions.
1. t < 5t + 24 t > –6
2. 5x – 9 ≤ 4.1x – 81 x ≤ –80
Hint: Clear the decimal!
3. 4b + 4(1 – b) > b – 9
Holt McDougal Algebra 1
b < 13
Inequalities with
3-5 Solving
Variables on Both Sides
Review Answers to Practice Handout!
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Homework
• Page 199 #s 1-12 all, 60
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
“Am I a solution?” activity!
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Wrap it up closure questions…
1)Complete my thought!
Whenever you multiply or divide by a negative
number, you must ….
2) How do I deal with 6>t?
3) Is 1000 a solution to h≥1000? What about
h>1000?
4) When do I use a open circle when graphing?
When do I use a closed circle?
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 2: Business Application
The Home Cleaning Company charges $312 to
power-wash the siding of a house plus $12 for
each window. Power Clean charges $36 per
window, and the price includes power-washing
the siding. How many windows must a house
have to make the total cost from The Home
Cleaning Company less expensive than Power
Clean?
Let w be the number of windows.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 2 Continued
Home
Cleaning
Company
siding
charge
312
plus
+
$12 per
window
12
times
# of
windows
is
less
than
Power
Clean
cost per
window
•
w
<
36
312 + 12w < 36w
– 12w –12w
312 < 24w
times
# of
windows.
•
w
To collect the variable terms,
subtract 12w from both sides.
Since w is multiplied by 24, divide
both sides by 24 to undo the
multiplication.
13 < w
The Home Cleaning Company is less expensive for
houses with more than 13 windows.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Check It Out! Example 2
A-Plus Advertising charges a fee of $24 plus
$0.10 per flyer to print and deliver flyers. Print
and More charges $0.25 per flyer. For how
many flyers is the cost at A-Plus Advertising
less than the cost of Print and More?
Let f represent the number of flyers printed.
A-Plus
Advertising plus
fee of $24
24
+
$0.10
per
flyer
times
0.10
•
Holt McDougal Algebra 1
Print and
# of
flyers
is less
than
More’s cost
f
<
0.25
times
# of
flyers.
per flyer
•
f
Inequalities with
3-5 Solving
Variables on Both Sides
Check It Out! Example 2 Continued
24 + 0.10f < 0.25f
–0.10f –0.10f
24
To collect the variable terms,
subtract 0.10f from both sides.
< 0.15f
Since f is multiplied by 0.15,
divide both sides by 0.15 to
undo the multiplication.
160 < f
More than 160 flyers must be delivered to make
A-Plus Advertising the lower cost company.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Some inequalities are true no matter what value is
substituted for the variable. For these inequalities,
all real numbers are solutions.
Some inequalities are false no matter what value is
substituted for the variable. These inequalities have
no solutions.
If both sides of an inequality are fully simplified
and the same variable term appears on both sides,
then the inequality has all real numbers as
solutions or it has no solutions. Look at the other
terms in the inequality to decide which is the case.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Additional Example 4A: All Real Numbers as
Solutions or No Solutions
Solve the inequality.
2x – 7 ≤ 5 + 2x
The same variable term (2x) appears on both
sides. Look at the other terms.
For any number 2x, subtracting 7 will always
result in a lower number than adding 5.
All values of x make the inequality true.
All real numbers are solutions.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Additional Example 4B: All Real Numbers as
Solutions or No Solutions
Solve the inequality.
2(3y – 2) – 4 ≥ 3(2y + 7)
6y – 8 ≥ 6y + 21
Distribute 2 on the left side
and 3 on the right side
and combine like terms.
The same variable term (6y) appears on both sides.
Look at the other terms.
For any number 6y, subtracting 8 will never
result in a higher number than adding 21.
No values of y make the inequality true.
There are no solutions.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Check It Out! Example 4a
Solve the inequality.
4(y – 1) ≥ 4y + 2
4y – 4 ≥ 4y + 2
Distribute 4 on the left side.
The same variable term (4y) appears on both sides.
Look at the other terms.
For any number 4y, subtracting 4 will never
result in a higher number than adding 2.
No values of y make the inequality true.
There are no solutions.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Check It Out! Example 4b
Solve the inequality.
x–2<x+1
The same variable term (x) appears on both
sides. Look at the other terms.
For any number x, subtracting 2 will always result
in a lesser number than adding 1.
All values of x make the inequality true.
All real numbers are solutions.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Lesson Quiz: Part II
4. Rick bought a photo printer and supplies for
$186.90, which will allow him to print photos
for $0.29 each. A photo store charges $0.55
to print each photo. How many photos must
Rick print before his total cost is less than
getting prints made at the photo store?
Rick must print more than 718 photos.
Holt McDougal Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Lesson Quiz: Part III
Solve each inequality.
5. 2y – 2 ≥ 2(y + 7)
no solutions
6. 2(–6r – 5) < –3(4r + 2)
all real numbers
Holt McDougal Algebra 1