2-1 Solving Linear Equations and Inequalities
Warm Up
Lesson Presentation
Lesson Quiz

2-1 Solving Linear Equations and Inequalities
Warm Up
Simplify each expression.
1. 2x + 5 – 3x –x + 5 2. –(w – 2) –w + 2
3. 6(2 – 3g) 12 – 18g
Graph on a number line
.
4. t > –2
–4 –3 –2 –1 0 1 2 3 4 5
5. Is 2 a solution of the inequality –2x < –6? Explain.
No; when 2 is substituted for x, the inequality is false:
–4 < –6
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
An equation is a mathematical statement that two expressions are equivalent. The solution set of an
equation is the value or values of the variable that make the equation true. A linear equation in one
variable can be written in the form ax = b, where a and b are constants and a ≠ 0.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Linear Equations in
One variable
4x = 8
3x – = –9
Nonlinear
Equations
+ 1 = 32
+ 1 = 41
2x – 5 = 0.1x +2 3 – 2 x = –5
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.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation. Do inverse operations in the reverse order of operations.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 1: Consumer Application
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?
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 1 Continued
Let m represent the number of additional minutes that
Nina used.
Model monthly charge plus additional minute charge times number of additional minutes
= total charge
12.95
+ 0.07
* m
= 14.56
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 1 Continued
Solve.
12.95 + 0.07m = 14.56
–12.95 –12.95
0.07m = 1.61
0.07
0.07
m = 23
Subtract 12.95 from both sides.
Divide both sides by 0.07.
Nina used 23 additional minutes.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 1
Stacked cups are to be placed in a pantry. One cup is 3.25 in. high and each additional cup raises the stack
0.25 in. How many cups fit between two shelves 14 in. apart?
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2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 1 Continued
Let c represent the number of additional cups needed.
Model one cup plus additional cup height times
3.25
+ 0.25
* number of additional cups c
=
= total height
14.00
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2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 1 Continued
Solve.
3.25 + 0.25c = 14.00
–3.25 –3.25
0.25c = 10.75
0.25
0.25
c = 43
Subtract 3.25 from both sides.
Divide both sides by 0.25.
44 cups fit between the 14 in. shelves.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 2: Solving Equations with the Distributive
Property
Solve 4(m + 12) = –36
Method 1
The quantity (m + 12) is multiplied by 4, so divide by 4 first.
4(m + 12) = –36
4 4
m + 12 = –9
–12 –12
m = –21
Divide both sides by 4.
Subtract 12 from both sides.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 2 Continued
Check 4(m + 12) = –36
4( –21 + 12) –36
4(–9) –36
–36 –36 
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 2 Continued
Solve 4(m + 12) = –36
Method 2
Distribute before solving.
4m + 48 = –36
–48 –48
4m = –84
4m –84
4 4
m = –21
Distribute 4.
Subtract 48 from both sides.
Divide both sides by 4.
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2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 2a
Solve 3(2 –3p) = 42.
Method 1
The quantity (2 – 3p) is multiplied by 3, so divide by 3 first.
3(2 – 3p) = 42
3 3
2 – 3p = 14
–2 –2
–3p = 12
–3 –3
p = –4
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Divide both sides by 3.
Subtract 2 from both sides.
Divide both sides by –3.

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 2a Continued
Check 3(2 – 3p) = 42
3(2 + 12 ) 42
6 + 36 42
42 42 
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2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 2a Continued
Solve 3(2 – 3p) = 42 .
Method 2
Distribute before solving.
6 – 9p = 42
–6 –6
–9p = 36
Distribute 3.
Subtract 6 from both sides.
Divide both sides by –9.
p = –4
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2-1 Solving Linear Equations and Inequalities
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Example 2b
Solve –3(5 – 4r) = –9.
Method 1
The quantity (5 – 4r) is multiplied by –3, so divide by
–3 first.
–3 –3
5 – 4r = 3
–5 –5
–4r = –2
Divide both sides by
–
3.
Subtract 5 from both sides.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 1
–4r –2
–4 –4 r =
Divide both sides by –4.
Check –3(5 –4r) = –9
–3(5 – 4 • ) –9
–3(5 – 2) –9
–3(3) –9
–9 –9 
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2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 2b Continued
Solve –3(5 – 4r) = –9.
Method 2
Distribute before solving.
–15 + 12r = –9
+15 +15
12r = 6
12r
=
6
12 12
r =
Distribute 3.
Add 15 to both sides.
Divide both sides by 12.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 3: Solving Equations with
Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12.
–11k + 25 = –6k – 10
+11k +11k
25 = 5k – 10
+10 + 10
35 = 5k
5 5
7 = k
Simplify each side by combining like terms.
Collect variables on the right side.
Add.
Collect constants on the left side.
Isolate the variable.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 3
Solve 3(w + 7) – 5w = w + 12.
–2w
+2w
+ 21 = w +
+2w
12
Simplify each side by combining like terms.
Collect variables on the right side.
21 = 3w + 12
–12 –12
9 = 3w
3 3
3 = w
Add.
Collect constants on the left side.
Isolate the variable.
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
Example 4A: Identifying Identities and Contractions
Solve 3v – 9 – 4v = –(5 + v).
3v – 9 – 4v = –(5 + v)
–9 – v = –5 – v
+ v + v
–9 ≠ –5 x
Simplify.
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
Example 4B: Identifying Identities and Contractions
Solve 2(x – 6) = –5x – 12 + 7x.
2(x – 6) = –5x – 12 + 7x
2x – 12 = 2x – 12
–2x –2x
Simplify.
–12 = –12  Identity
The solutions set is all real number, or  .
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 4a
Solve 5(x – 6) = 3x – 18 + 2x.
5(x – 6) = 3x – 18 + 2x
5x – 30 = 5x – 18
–5x –5x
–30 ≠ –18 x
Simplify.
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
Check It Out!
Example 4b
Solve 3(2 –3x) = –7x – 2(x –3).
3(2 –3x) = –7x – 2(x –3)
6 – 9x = –9x + 6
+ 9x +9x
6 = 6 
Simplify.
Identity
The solutions set is all real numbers, or  .
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠.
The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality.
The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must
reverse the inequality symbol.
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
Helpful
Hint
To check an inequality, test
• the value being compared with x
• a value less than that, and
• a value greater than that.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Example 5: Solving Inequalities
Solve and graph 8a –2 ≥ 13a + 8.
8a – 2 ≥ 13a + 8
–13a –13a
–5a – 2 ≥ 8
+2 +2
–5a ≥ 10
–5a ≤ 10
–5 –5
a ≤ –2
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Subtract 13a from both sides.
Add 2 to both sides.
Divide both sides by –5 and reverse the inequality.

2-1 Solving Linear Equations and Inequalities
Example 5 Continued
Solve and graph 8a – 2 ≥ 13a + 8.
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.
–18 ≥ –18
So –2 is a solution.
 –10 ≥ –5 x
So –1 is not a solution.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 5
Solve and graph x + 8 ≥ 4x + 17.
x + 8 ≥ 4x + 17
–x –x
8 ≥ 3x +17
–17 –17
–9 ≥ 3x
–9 ≥ 3x
3 3
–3 ≥ x or x ≤ –3
Subtract x from both sides.
Subtract 17 from both sides.
Divide both sides by 3.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Check It Out!
Example 5 Continued
Solve and graph x + 8 ≥ 4x + 17.
Check Test values in the original inequality.
•
–6 –5 –4 –3 –2 –1 0 1 2 3
Test x = –6 Test x = –3 Test x = 0
–6 + 8 ≥ 4( –6 ) + 17
–3
+8 ≥ 4( –3 ) + 17 0 + 8 ≥ 4( 0 ) + 17
2 ≥ –7 
So –6 is a solution.
5 ≥ 5 
So –3 is a solution.
8 ≥ 17 x
So 0 is not a solution.
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part I
1. Alex pays $19.99 for cable service each month.
He also pays $2.50 for each movie he orders through the cable company’s pay-per-view service. If his bill last month was $32.49, how many movies did Alex order?
5 movies
Holt Algebra 2

2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part II
Solve.
2. 2(3x – 1) = 34
x = 6
3. 4y – 9 – 6y = 2(y + 5) – 3 y = –4
4. r + 8 – 5r = 2(4 – 2r) all real numbers, or 
5. –4(2m + 7) = (6 – 16m) no solution, or
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2-1 Solving Linear Equations and Inequalities
Lesson Quiz: Part III
5. Solve and graph.
12 + 3q > 9q – 18 q < 5
°
–2 –1 0 1 2 3 4 5 6 7
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