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Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Section 2.1
Solving Linear Equations and Inequalities
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
Homework• Pg 94 #22 - 39
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
These properties also apply to inequalities expressed with >, ≥, and ≤.
Holt Algebra 2
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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
2-1 Solving Linear Equations and Inequalities
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