holt algebra 2 2-1 solving linear equations and inequalities section 2.1 solving linear equations...

Holt Algebra 2

2-1 Solving Linear Equations and Inequalities

Section 2.1

Solving Linear Equations and Inequalities

Holt Algebra 2


Homework• Pg 94 #22 - 39

Holt Algebra 2


Linear Equations in One variable

Nonlinear Equations

4x = 8

3x – = –9

2x – 5 = 0.1x +2

Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator.

Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality.

+ 1 = 32

+ 1 = 41

3 – 2x = –5

Holt Algebra 2


Holt Algebra 2


The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use?

Consumer Application

Holt Algebra 2


Solving Equations with the Distributive Property

Solve 4(m + 12) = –36

Divide both sides by 4.

The quantity (m + 12) is multiplied by 4, so divide by 4 first.

4(m + 12) = –364 4

m + 12 = –9

m = –21

–12 –12 Subtract 12 from both sides.

Holt Algebra 2


Divide both sides by 3.

The quantity (2 – 3p) is multiplied by 3, so divide by 3 first.

3(2 – 3p) = 42

3 3

Solve 3(2 –3p) = 42.

Subtract 2 from both sides.

–3p = 12

2 – 3p = 14 –2 –2

–3 –3 Divide both sides by –3.

p = –4

Holt Algebra 2


Solving Equations with Variables on Both Sides

Simplify each side by combining like terms.

–11k + 25 = –6k – 10

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+11k +11k

25 = 5k – 10

35 = 5k

5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12.

Holt Algebra 2


You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution.

An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.

Holt Algebra 2


Solve 3v – 9 – 4v = –(5 + v).

Identifying Identities and Contractions

3v – 9 – 4v = –(5 + v)

Simplify.–9 – v = –5 – v + v + v

–9 ≠ –5 x Contradiction

The equation has no solution. The solution set is the empty set, which is represented by the symbol .

Holt Algebra 2


Solve 2(x – 6) = –5x – 12 + 7x.

: Identifying Identities and Contractions

2(x – 6) = –5x – 12 + 7x Simplify.2x – 12 = 2x – 12

–2x –2x

–12 = –12 Identity

The solutions set is all real number, or .

Holt Algebra 2


These properties also apply to inequalities expressed with >, ≥, and ≤.

Holt Algebra 2


Solve and graph 8a –2 ≥ 13a + 8.

Solving Inequalities

Subtract 13a from both sides.8a – 2 ≥ 13a + 8

–13a –13a

–5a – 2 ≥ 8Add 2 to both sides. +2 +2

–5a ≥ 10Divide both sides by –5 and reverse the inequality.

–5 –5

–5a ≤ 10

a ≤ –2

Holt Algebra 2


Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

•

Test x = –4 Test x = –2 Test x = –1

8(–4) – 2 ≥ 13(–4) + 8 8(–2) – 2 ≥ 13(–2) + 8 8(–1) – 2 ≥ 13(–1) + 8

–34 ≥ –44

So –4 is a solution.

So –1 is not a solution.

So –2 is a solution.

–18 ≥ –18 –10 ≥ –5 x

Solve and graph 8a – 2 ≥ 13a + 8.

Holt Algebra 2


Solve and graph x + 8 ≥ 4x + 17.

Subtract x from both sides.x + 8 ≥ 4x + 17

–x –x

8 ≥ 3x +17Subtract 17 from both sides.–17 –17

–9 ≥ 3xDivide both sides by 3.

3 3

–9 ≥ 3x

–3 ≥ x or x ≤ –3

holt algebra 2 2-1 solving linear equations and inequalities section 2.1 solving linear equations...

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