Solving Linear Inequalities

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Warm-up

-4 < x ≤ 6

x ≤ -4 or x>6

-9

-8

-7

-6

-5

-4

-3

-2

-1

0 1 2 3 4 5 6 7 8 9

-9

-8

-7

-6

-5

-4

-3

-2

-1

0 1 2 3 4 5 6 7 8 9

-1 < x ≤ 9 -9

-8

-7

-6

-5

-4

-3

-2

-1

0 1 2 3 4 5 6 7 8 9

x ≤ 1 or x>7 -9

-8

-7

-6

-5

-4

-3

-2

-1

0 1 2 3 4 5 6 7 8 9

Solving Compound Inequalities

-8 < 2x ≤ 102 2

2-4 < x ≤ 5

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Example 1

This compound inequality is a conjunction because it uses the word “and”.

x is less than –4 and greater than 5.

2-step Compound Inequalities

2x-1 < -9 or 2x-1 ≥ 5

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Example 2

2x-1 < -9 +1 +1

2x < -822

x < -4

2x-1 ≥ 5+1 +1

2x ≥ 622

x ≥ 3x > -4 or x

≥ 9

This compound inequality is a disjunction because it uses the word “or”.

Your Turn

2x < -12 or 2x ≥ 10-11 < 2x+1 ≤ 7

3 < 3x < 9 2x-3 < -15 or 2x-3 ≥ 7

-13 ≤ 5x-2 < 18 4x < -32 or 4x ≥ 40

-6 < x ≤ 3 x<-6 or x ≥ 5

1 < x < 3 x<-6 or x ≥ 5

-11/5 ≤ x < 4 x<-8 or x ≥ 10

A quadratic inequality in one variable is an inequality which can be written in the form

ax2 + bx + c > 0 (a ≠ 0)

The symbols ≥, , and ≤ may also be used.

for a, b, c real numbers.

A solution of a quadratic inequality in one variable is a number which, when substituted for the variable, results in a true inequality.

Example: Which of the values of x are solutions of x2 + 3x − 4 ≤ 0 ?

− 1 (−1)2 + 3(−1) – 4 − 6 ≤ 0 true yes

0 (0)2 + 3(0) – 4 − 4 ≤ 0 true yes

2 (2)2 + 3(2) – 4 6 ≤ 0 false no

3 (3)2 + 3(3) – 4 14 ≤ 0 false no

x x2 + 3x – 4 x2 + 3x – 4 ≤ 0 Solution?

Find the solution set of x2 + 3x − 4 ≤ 0. x2 + 3x − 4 ≤ 0

(x + 4)(x - 1) ≤ 0

(x +4) ≤ 0 (x - 1) ≤ 0 x ≤ -4 x ≤ 1x ≤ -4 and x ≤ 1 is a conjunction, therefore

-4 ≤x ≤ 1

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Solve and graph the solution set of x2 − 6x + 5 < 0.

YOUR TURN!

1 < x < 5

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Example: Solve and graph the solution set of x2 − x ≥ 6. x2 - x − 6 ≥ 0

(x + 2)(x - 3) ≥ 0

(x +2) ≤ 0 (x - 3) ≥ 0 x ≤ -2 x ≥ 3This inequality is a disjunction, therefore

x ≤ -2 or x ≥ 3

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9

Solve and graph the solution set of x2 − 7x + 12 > 0.YOUR TURN!

x<3 or x>4

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9