Transcript 3-5 - Ms. Muehleck`s Math Class Website

Inequalities with
3-5 Solving
Variables on Both Sides
Lesson Objective: I will be able to …
• Solve inequalities in one variable that contain
variable terms on both sides
Language Objective: I will be able to …
• Read, write, and listen about vocabulary, key
concepts, and examples
Holt Algebra 1
Inequalities with
3-5 Solving
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.
Page 32
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 Algebra 1
5
6
Inequalities with
3-5 Solving
Variables on Both Sides
Page 32
Your Turn 1
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 Algebra 1
0
2
4
6
8 10
Inequalities with
3-5 Solving
Variables on Both Sides
Example 2: Business Application
Page 33
The Home Cleaning Company charges $312 to powerwash 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.
Home
Cleaning
Company
siding
charge
312
plus
+
Holt Algebra 1
$12 per
window
12
times
# of
windows
is
less
than
Power
Clean
cost per
window
•
w
<
36
times
# of
windows.
•
w
Inequalities with
3-5 Solving
Variables on Both Sides
Example 2 Continued
312 + 12w < 36w
– 12w –12w
312 < 24w
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 3: Simplify Each Side Before
Solving
Solve the inequality and graph the solutions.
Page 34
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
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.
–6 > 3 + k
–3 –3
–9 > k
–12
Holt Algebra 1
–9
–6
–3
0
3
Inequalities with
3-5 Solving
Variables on Both Sides
Page 34
Your Turn 3
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 the right
5(2 – r) ≥ 3(r – 2)
side of the inequality.
5(2) – 5(r) ≥ 3(r) + 3(–2)
10 – 5r ≥ 3r – 6
+6
+6
16 − 5r ≥ 3r
+ 5r +5r
16
8
≥ 8r
8
2≥r
Holt Algebra 1
Since 6 is subtracted from 3r, add 6
to both sides to undo the subtraction.
Since 5r is subtracted from 16 add 5r
to both sides to undo the subtraction.
Since r is multiplied by 8, divide both
sides by 8 to undo the multiplication.
–6
–4
–2
0
2
4
Inequalities with
3-5 Solving
Variables on Both Sides
Page 31
Holt Algebra 1
Inequalities with
3-5 Solving
Variables on Both Sides
Example 4: Identities and
Contradictions
Solve the inequality.
Page 34
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 5: Identities and
Contradictions
Solve the inequality.
Page 35
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
Page 35
Your Turn 5
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
Homework
Assignment #6
•
3-5 #4-6, 14, 16
Holt Algebra 1