§ 2.8
DESCRIPTION
Solving Linear Inequalities. § 2.8. Linear Inequalities in One Variable. A linear inequality in one variable is an inequality that can be written in the form ax + b < c where a , b , and c are real numbers and a is not 0. - PowerPoint PPT PresentationTRANSCRIPT
§ 2.8
Solving Linear Inequalities
Martin-Gay, Beginning and Intermediate Algebra, 4ed 22
Linear Inequalities in One Variable
A linear inequality in one variable is an inequality that can be written in the form
ax + b < c
where a, b, and c are real numbers and a is not 0.
This definition and all other definitions, properties and steps in this section also hold true for the inequality symbols >, , or .
Martin-Gay, Beginning and Intermediate Algebra, 4ed 33
Graphing Solutions to Linear Inequalities• Use a number line.• Use a closed circle at the endpoint of an interval if
you want to include the point.• Use an open circle at the endpoint if you DO NOT
want to include the point.
7Represents the set {xx 7}.
-4Represents the set {xx > – 4}.
Solutions to Linear Inequalities
Interval notation, is used to write solution sets of inequalities.• Use a parenthesis if you want to include the number• Use a bracket if you DO NOT want to include the number..
Martin-Gay, Beginning and Intermediate Algebra, 4ed 44
Solutions to Linear Inequalitiesx < 3
-5 -4 -3 -2 -1 0 1 2 3 4 5
–1.5 x 3
–2 < x < 0-5 -4 -3 -2 -1 0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval notation: (–∞, 3)
Interval notation: (–2, 0)
Interval notation: [–1.5, 3)
NOT included
Included
Martin-Gay, Beginning and Intermediate Algebra, 4ed 55
Addition Property of InequalityIf a, b, and c are real numbers, then
a < b and a + c < b + c are equivalent inequalities.
Multiplication Property of Inequality1. If a, b, and c are real numbers, and c is positive, then
a < b and ac < bc are equivalent inequalities.
2. If a, b, and c are real numbers, and c is negative, thena < b and ac > bc are equivalent inequalities.
Properties of Inequality
Martin-Gay, Beginning and Intermediate Algebra, 4ed 66
Solving Linear Inequalities in One Variable1) Clear the inequality of fractions by multiplying both sides
by the LCD of all fractions of the inequality.2) Remove grouping symbols by using the distributive
property.3) Simplify each side of equation by combining like terms.4) Write the inequality with variable terms on one side and
numbers on the other side by using the addition property of equality.
5) Get the variable alone by using the multiplication property of equality.
Solving Linear Inequalities
Martin-Gay, Beginning and Intermediate Algebra, 4ed 77
Solving Linear Inequalities
Example: Solve the inequality. Graph the solution and give your answer in interval notation.
2(x – 3) < 4x + 10
2x – 6 < 4x + 10 Distribute.
– 6 < 2x + 10 Subtract 2x from both sides.
– 16 < 2x Subtract 10 from both sides.
– 8 < x or x > – 8 Divide both sides by 2.
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 (–8, ∞)
Martin-Gay, Beginning and Intermediate Algebra, 4ed 88
Since 0 is always greater than –7, the solution is all real numbers. (Any value we put in for x in the original statement will give us a true inequality.)
Solving Linear InequalitiesExample: Solve the inequality. Graph the solution
and give your answer in interval notation.x + 5 x – 2
x x – 7 Subtract 5 from both sides.
0 – 7 Subtract x from both sides.
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4
(Always true!)
(–∞ ∞)
Martin-Gay, Beginning and Intermediate Algebra, 4ed 99
Solving Linear InequalitiesExample: Solve the inequality. Give your answer in
interval notation.a.) 9 < z + 5 < 13
b.) –7 < 2p – 3 ≤ 5a.) 9 < z + 5 < 13 4 < z < 8 Subtract 5 from all three parts.
(4, 8)
b.) –7 < 2p – 3 ≤ 5 –4 < 2p ≤ 8 Add 3 to all three parts. –2 < p ≤ 4 Divide all three parts by 2.
(–2, 4]
Martin-Gay, Beginning and Intermediate Algebra, 4ed 1010
3x + 9 5(x – 1)
3x + 9 5x – 5 Use distributive property on right side.
3x – 3x + 9 5x – 3x – 5 Subtract 3x from both sides.
9 2x – 5 Simplify both sides.
14 2x Simplify both sides.
7 x Divide both sides by 2.
9 + 5 2x – 5 + 5 Add 5 to both sides.
Solution:7]
Solving Linear Inequalities
Example: Solve the inequality. Give your answer in graph form.
Martin-Gay, Beginning and Intermediate Algebra, 4ed 1111
A compound inequality contains two inequality symbols.
To solve the compound inequality, perform operations simultaneously to all three parts of the inequality (left, middle and right).
Compound Inequalities
0 4(5 – x) < 8
This means 0 4(5 – x) and 4(5 – x) < 8.
Martin-Gay, Beginning and Intermediate Algebra, 4ed 1212
The solution is (3,5].
– 20 – 4x < – 12 Simplify each part. 5 x > 3 Divide each part by –4.Remember that the sign changes direction when you divide by a negative number.
0 – 20 20 – 20 – 4x < 8 – 20 Subtract 20 from each part.
Solving Compound Inequalities
Example: Solve the inequality. Give your answer in interval notation.
0 20 – 4x < 8 Use the distributive property.
0 20 – 4x < 8
Martin-Gay, Beginning and Intermediate Algebra, 4ed 1313
Inequality Applications
Example: Six times a number, decreased by 2, is at least 10. Find the number.
1.) UNDERSTAND
Let x = the unknown number.
“Six times a number” translates to 6x,
“decreased by 2” translates to 6x – 2,
“is at least 10” translates ≥ 10.
Continued
Martin-Gay, Beginning and Intermediate Algebra, 4ed 1414
decreased
–
Six times a number
6x
by 2
2
is at least
≥
10
10
Finding an Unknown Number
Example continued:
2.) TRANSLATE
Continued
Martin-Gay, Beginning and Intermediate Algebra, 4ed 1515
Finding an Unknown Number
Example continued:3.) SOLVE
6x – 2 ≥ 106x ≥ 12 Add 2 to both sides.x ≥ 2 Divide both sides by 6.
4.) INTERPRETCheck: Replace “number” in the original statement of the problem with a number that is 2 or greater. Six times 2, decreased by 2, is at least 10
6(2) – 2 ≥ 10 10 ≥ 10 State: The number is 2.