Transcript Solve the inequality.

5.2
Solving Inequalities with
Variables on Both Sides
Example 1: 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.
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Holt McDougal Algebra 1
5
6
5.2
Solving Inequalities with
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
Solving Inequalities with
Variables on Both Sides
5.2
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
5.2
Solving Inequalities with
Variables on Both Sides
Check It Out! Example 3
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
5.2
Solving Inequalities with
Variables on Both Sides
Check It Out! Example 3 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
5.2
Solving Inequalities with
Variables on Both Sides
Check It Out! Example 4
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
5.2
Solving Inequalities with
Variables on Both Sides
Check It Out! Example 4 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
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5.2
Solving Inequalities with
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
5.2
Solving Inequalities with
Variables on Both Sides
Additional Example 5: 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
5.2
Solving Inequalities with
Variables on Both Sides
Additional Example 6: 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