Transcript 3, divide both sides by

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
Holt Algebra 1
–5
–4
–3
–2
–1
0
Inequalities with
3-5 Solving
Variables on Both Sides
Objective
Solve inequalities that contain variable
terms on both sides.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Some inequalities have variable terms on both
sides of the inequality symbol. You can solve
these inequalities 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 Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Directions:
Solve the inequality and graph the solutions.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 1
y ≤ 4y + 18
y ≤ 4y + 18
–y –y
To collect the variable terms on one
side, subtract y from both sides.
0 ≤ 3y + 18
–18
– 18
Since 18 is added to 3y, subtract 18
from both sides to undo the
addition.
–18 ≤ 3y
Since y is multiplied by 3, divide both
sides by 3 to undo the
multiplication.
–6 ≤ y (or y  –6)
Holt Algebra 1
–10 –8 –6 –4 –2
0
2
4
6
8 10
Inequalities with
3-5 Solving
Variables on Both Sides
Example 2
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 Algebra 1
5
6
Inequalities with
3-5 Solving
Variables on Both Sides
Example 3
4x ≥ 7x + 6
4x ≥ 7x + 6
–7x –7x
To collect the variable terms on one
side, subtract 7x from both sides.
–3x ≥ 6
x ≤ –2
Since x is multiplied by –3, divide
both sides by –3 to undo the
multiplication. Change ≥ to ≤.
–10 –8 –6 –4 –2
Holt Algebra 1
0
2
4
6
8 10
Inequalities with
3-5 Solving
Variables on Both Sides
Example 4
5t + 1 < –2t – 6
5t + 1 < –2t – 6
+2t
+2t
7t + 1 < –6
– 1 < –1
7t
< –7
7t < –7
7
7
t < –1
–5 –4 –3 –2 –1
Holt Algebra 1
0
1
2
To collect the variable terms on
one side, add 2t to both sides.
Since 1 is added to 7t, subtract 1
from both sides to undo the
addition.
Since t is multiplied by 7, divide
both sides by 7 to undo the
multiplication.
3
4
5
Inequalities with
3-5 Solving
Variables on Both Sides
Example 5: 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 Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 5 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 Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 6
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
+
Holt Algebra 1
$0.10
per
flyer
times
0.10
•
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
Example 6 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 Algebra 1
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 Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 7
2(k – 3) > 6 + 3k – 3
Distribute 2 on the left side of
2(k – 3) > 3 + 3k
the inequality.
2k + 2(–3) > 3 + 3k
2k – 6 > 3 + 3k
–2k
– 2k
–6 > 3 + k
–3 –3
–9 > k
Holt Algebra 1
To collect the variable terms,
subtract 2k from both
sides.
Since 3 is added to k, subtract 3
from both sides to undo the
addition.
Inequalities with
3-5 Solving
Variables on Both Sides
Example 7 Continued
–9 > k
–12
Holt Algebra 1
–9
–6
–3
0
3
Inequalities with
3-5 Solving
Variables on Both Sides
Example 8
0.9y ≥ 0.4y – 0.5
0.9y ≥ 0.4y – 0.5
–0.4y –0.4y
To collect the variable terms,
subtract 0.4y from both sides.
0.5y ≥
– 0.5
0.5y ≥ –
0.5
0.5
0.5
y ≥ –1
–5 –4 –3 –2 –1
Holt Algebra 1
0
1
2
3
Since y is multiplied by 0.5,
divide both sides by 0.5 to
undo the multiplication.
4
5
Inequalities with
3-5 Solving
Variables on Both Sides
Example 9
5(2 – r) ≥ 3(r – 2)
Distribute 5 on the left side of the
inequality and distribute 3 on
5(2 – r) ≥ 3(r – 2)
the right side of the inequality.
5(2) – 5(r) ≥ 3(r) + 3(–2)
Since 6 is subtracted from 3r,
10 – 5r ≥ 3r – 6
add 6 to both sides to undo
+6
+6
the subtraction.
16 − 5r ≥ 3r
Since 5r is subtracted from 16
+ 5r +5r
add 5r to both sides to undo
the subtraction.
16
≥ 8r
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 9 Continued
16 ≥ 8r
Since r is multiplied by 8, divide
both sides by 8 to undo the
multiplication.
2≥r
–6
Holt Algebra 1
–4
–2
0
2
4
Inequalities with
3-5 Solving
Variables on Both Sides
Example 10
0.5x – 0.3 + 1.9x < 0.3x + 6
2.4x – 0.3 < 0.3x + 6
2.4x – 0.3 < 0.3x + 6
+ 0.3
+ 0.3
2.4x
–0.3x
2.1x
< 0.3x + 6.3
–0.3x
<
6.3
Simplify.
Since 0.3 is subtracted
from 2.4x, add 0.3 to
both sides.
Since 0.3x is added to
6.3, subtract 0.3x from
both sides.
Since x is multiplied by
2.1, divide both sides
by 2.1.
x<3
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 10 Continued
x<3
–5 –4 –3 –2 –1
Holt Algebra 1
0
1
2
3
4
5
Inequalities with
3-5 Solving
Variables on Both Sides
There are special cases of inequalities called
identities and contradictions.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 11
2x – 7 ≤ 5 + 2x
2x – 7 ≤ 5 + 2x
–2x
–2x
–7 ≤ 5
Subtract 2x from both sides.
True statement.
The inequality 2x − 7 ≤ 5 + 2x is an identity. All
values of x make the inequality true. Therefore,
all real numbers are solutions.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 12
2(3y – 2) – 4 ≥ 3(2y + 7)
Distribute 2 on the left side
and 3 on the right side.
2(3y) – 2(2) – 4 ≥ 3(2y) + 3(7)
2(3y – 2) – 4 ≥ 3(2y + 7)
6y – 4 – 4 ≥ 6y + 21
6y – 8 ≥ 6y + 21
–6y
–6y
Subtract 6y from both sides.
False statement.
–8 ≥ 21
No values of y make the inequality true.
There are no solutions.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 13
Solve the inequality.
4(y – 1) ≥ 4y + 2
4(y – 1) ≥ 4y + 2
Distribute 4 on the left side.
4(y) + 4(–1) ≥ 4y + 2
4y – 4 ≥ 4y + 2
–4y
–4y
–4 ≥ 2 
Subtract 4y from both sides.
False statement.
No values of y make the inequality true.
There are no solutions.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 14
Solve the inequality.
x–2<x+1
x–2<x+1
–x
–x
–2 < 1
Subtract x from both sides.
True statement.
All values of x make the inequality true.
All real numbers are solutions.
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Lesson Summary: Part I
Solve each inequality and graph the solutions.
1. t < 5t + 24 t > –6
2. 5x – 9 ≤ 4.1x – 81 x ≤ –80
3. 4b + 4(1 – b) > b – 9
Holt Algebra 1
b < 13
Inequalities with
3-5 Solving
Variables on Both Sides
Lesson Summary: 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 Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Lesson Summary: Part III
Solve each inequality.
5. 2y – 2 ≥ 2(y + 7)
contradiction, no solution
6. 2(–6r – 5) < –3(4r + 2)
identity, all real numbers
Holt Algebra 1