Transcript Solve the inequality.

Inequalities with
2-5 Solving
Variables on Both Sides
Objective
Solve inequalities that contain variable
terms on both sides.
Holt McDougal Algebra 1
Inequalities with
2-5 Solving
Variables on Both Sides
Example 1A: Solving Inequalities with Variables on
Both Sides
Solve the inequality and graph the solutions.
y ≤ 4y + 18
To collect the variable terms on one
y ≤ 4y + 18
side, subtract y from both sides.
–y –y
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 McDougal Algebra 1
–10 –8 –6 –4 –2
0
2
4
6
8 10
Inequalities with
2-5 Solving
Variables on Both Sides
Example 1B: Solving Inequalities with Variables on
Both Sides
Solve the inequality and graph the solutions.
4m – 3 < 2m + 6
To collect the variable terms on one
–2m
– 2m
side, subtract 2m from both sides.
2m – 3 <
+6
+3
+3
2m
<
9
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
2-5 Solving
Variables on Both Sides
Check It Out! Example 1a
Solve the inequality and graph the solutions.
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 McDougal Algebra 1
0
2
4
6
8 10
Inequalities with
2-5 Solving
Variables on Both Sides
Check It Out! Example 1b
Solve the inequality and graph the solutions.
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 McDougal 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
2-5 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
2-5 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
2-5 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
2-5 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
2-5 Solving
Variables on Both Sides
Example 3A: Simplify Each Side Before Solving
Solve the inequality and graph the solutions.
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 McDougal 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
2-5 Solving
Variables on Both Sides
Example 3A Continued
–9 > k
–12
–9
Holt McDougal Algebra 1
–6
–3
0
3
Inequalities with
2-5 Solving
Variables on Both Sides
Example 3B: Simplify Each Side Before Solving
Solve the inequality and graph the solution.
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 McDougal 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
2-5 Solving
Variables on Both Sides
Check It Out! Example 3a
Solve the inequality and graph the solutions.
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 McDougal Algebra 1
Inequalities with
2-5 Solving
Variables on Both Sides
Check It Out! Example 3a Continued
16 ≥ 8r
Since r is multiplied by 8, divide
both sides by 8 to undo the
multiplication.
2≥r
–6
–4
Holt McDougal Algebra 1
–2
0
2
4
Inequalities with
2-5 Solving
Variables on Both Sides
Check It Out! Example 3b
Solve the inequality and graph the solutions.
0.5x – 0.3 + 1.9x < 0.3x + 6
Simplify.
2.4x – 0.3 < 0.3x + 6
Since 0.3 is subtracted
2.4x – 0.3 < 0.3x + 6
from 2.4x, add 0.3 to
+ 0.3
+ 0.3
both sides.
2.4x
< 0.3x + 6.3
Since 0.3x is added to
–0.3x
–0.3x
6.3, subtract 0.3x from
both sides.
2.1x
<
6.3
Since x is multiplied by
2.1, divide both sides
by 2.1.
x<3
Holt McDougal Algebra 1
Inequalities with
2-5 Solving
Variables on Both Sides
Check It Out! Example 3b Continued
x<3
–5 –4 –3 –2 –1
Holt McDougal Algebra 1
0
1
2
3
4
5
Inequalities with
2-5 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
2-5 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
2-5 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
2-5 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