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Chapter 1
Section 4
Solving Inequalities
Solving Inequalities
ALGEBRA 2 LESSON 1-4
(For help, go to Lessons 1-1 and 1-3.)
State whether each inequality is true or false.
1. 5 < 12
2. 5 < –12
3. 5 >
– 12
4. 5 <
– –12
5. 5 <
–5
6. 5 >
– 5
Solve each equation.
7. 3x + 3 = 2x – 3
8. 5x = 9(x – 8) + 12
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Solutions
1. 5 < 12, true
2. 5 < –12, false
3. 5 <
– 12, false
4. 5 >
– –12, false
5. 5 >
– 5, true
6. 5 <
– 5, true
7. 3x + 3 = 2x – 3
8. 5x = 9(x – 8) + 12
3x – 2x = –3 – 3
x = –6
5x = 9x – 72 + 12
–4x = –60
x = 15
Inequalities
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The solutions include more than one number
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All of the rules for solving equations apply to
inequalities, with one added:
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Ex: 2 < x ;values that x could be include 3, 7, 45…
If you multiply or divide by a NEGATIVE you must FLIP the
sign. (< becomes > and > becomes <)
When graphing on a number line:

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Open dot for < or >
Closed (solid) dot for ≤ or ≥
The shading should be easy to see (a slightly elevated line is
ok) --- see examples
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Solve –2x < 3(x – 5). Graph the solution.
–2x < 3(x – 5)
–2x < 3x – 15 Distributive Property
–5x < –15
x >3
Subtract 3x from both sides.
Divide each side by –5 and reverse
the inequality.
Try These Problems
Solve each inequality. Graph the solution.
a)
3x – 6 < 27
a)
3x < 33
x < 11
12 ≥ 2(3n + 1) + 22
b)
a)
12 ≥ 6n + 2 + 22
12 ≥ 6n + 24
-12 ≥ 6n
-2 ≥ n
Solve 7x ≥ 7(2 + x). Graph the solution.
7x ≥ 7(2 + x)
7x ≥ 14 + 7x
0 ≥ 14
Distributive Property
Subtract 7x from both sides.
0
The last inequality is always false, so 7x ≥ 7(2 + x) is
always false. It has no solution.
Try These Problems
Solve. Graph the solution.
a)
2x < 2(x + 1) + 3
a)
b)
2x < 2x + 2 + 3
2x < 2x + 5
0<5
All Real Numbers
4(x – 3) + 7 ≥ 4x + 1
a)
0
4x – 12 + 7 ≥ 4x + 1
4x + 12 ≥ 4x + 1
12 ≥ 1
No Solution
0
Solving Inequalities
A real estate agent earns a salary of $2000 per month plus
4% of the sales. What must the sales be if the salesperson is to have
a monthly income of at least $5000?
Relate: $2000 + 4% of sales >
– $5000
Define: Let x = sales (in dollars).
Write:
2000 + 0.04x >
– 5000
0.04x >
– 3000
x >
– 75,000
Subtract 2000 from each side.
Divide each side by 0.04.
The sales must be greater than or equal to $75,000.
Try This Problem
A salesperson earns a salary of $700 per month plus
2% of the sales. What must the sales be if the
salesperson is to have a monthly income of at least
$1800?
700 + .02x ≥ 1800
.02x ≥ 1100
x ≥ 55000
The sales must be at least $55,000.
Compound Inequalities
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Compound Inequality – a pair of inequalities joined
by and or or
Ex: -1 < x and x ≤ 3 which can be written as
-1 < x ≤ 3
x < -1 or x ≥ 3
For and statements the value must satisfy both
inequalities
For or statements the value must satisfy one of the
inequalities
And Inequalities
a)
Graph the solution of
3x – 1 > -28 and 2x + 7 < 19.
3x > -27 and 2x < 12
x > -9 and x < 6
b)
Graph the solution of -8 < 3x + 1 <19
-9 < 3x < 18
-3 < x < 6
Or Inequalities
ALGEBRA 2 LESSON 1-4
Graph the solution of 3x + 9 < –3 or –2x + 1 < 5.
3x + 9 < –3
or
–2x + 1 < 5
3x < –12
–2x < 4
x < –4 or x > –2
Try These Problems
Graph the solution of 2x > x + 6 and x – 7 < 2
a)
a)
x > 6 and x < 9
Graph the solution of x – 1 < 3 or x + 3 > 8
b)
a)
x < 4 or x > 11
Homework
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Practice 1.4 All omit 23