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 is an equation 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
2x – 5 = 0.1x +2
Nonlinear
Equations
+ 1 = 32
+ 1 = 41
3 – 2x = –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
Properties of Equality
Addition / Subtraction Property of Equality
If you add or subtract the same quantity to or
from both sides of an equation, the equation
will still be true.
Example:
3 3
3 2  3 2
Holt Algebra 2
3 3
3 4  3 4
2-1
Solving Linear Equations and Inequalities
Properties of Equality
Multiplication / Division Property of Equality
If you multiply or divide both sides of an equation
by the same quantity, the equation will still be
true.
Example:
3 3
3(2)  3(2)
Holt Algebra 2
3 3
3 3

4 4
2-1
Solving Linear Equations and Inequalities
Solving Linear Equations
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.
Holt Algebra 2
2-1
Solving Linear Equations and Inequalities
Example
Method 1
Distribute before solving.
4(m + 12) = –36
4m + 48 = –36
–48 –48
Distribute 4.
Subtract 48 from both sides.
4m = –84
4m –84
=
4
4
m = –21
Holt Algebra 2
Divide both sides by 4.
2-1
Solving Linear Equations and Inequalities
Example
Solve 4(m + 12) = –36
Method 2
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
Holt Algebra 2
Divide both sides by 4.
Subtract 12 from both sides.
2-1
Solving Linear Equations and Inequalities
Example 1
Method 1
Distribute before solving.
–3(5 – 4r) = –9.
–15 + 12r = –9
+15
+15
12r = 6
12r
6
=
12
12
r=
Holt Algebra 2
Distribute 3.
Add 15 to both sides.
Divide both sides by 12.
2-1
Solving Linear Equations and Inequalities
Example 1
Solve –3(5 – 4r) = –9.
Method 2
–3(5 – 4r) = –9
–3
–3
5 – 4r = 3
–5
–5
Divide both sides by –3.
Subtract 5 from both sides.
–4r = –2
=
–4r
–4
r=
Holt Algebra 2
–2
–4
Divide both sides by –4.
2-1
Solving Linear Equations and Inequalities
Method 1
Example 2
Distribute before solving.
3(2 – 3p) = 42
6 – 9p = 42
–6
–6
–9p = 36
–9p 36
=
–9 –9
p = –4
Holt Algebra 2
Distribute 3.
Subtract 6 from both sides.
Divide both sides by –9.
2-1
Solving Linear Equations and Inequalities
Example 2
Solve 3(2 –3p) = 42.
Method 2
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
Holt Algebra 2
Divide both sides by 3.
Subtract 2 from both sides.
Divide both sides by –3.
2-1
Solving Linear Equations and Inequalities
Example 3: Solving Equations with
Variables on Both Sides
Solve 3k– 14k + 25 = 2 – 6k – 12.
Simplify each side by combining
–11k + 25 = –6k – 10
like terms.
+11k
+11k
Collect variables on the right side.
25 = 5k – 10
+10
+ 10
35 = 5k
5
5
7=k
Holt Algebra 2
Add.
Collect constants on the left side.
Isolate the variable.
2-1
Solving Linear Equations and Inequalities
A contradiction is an equation that has no solutions.
Example: 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
An identity is an equation that is true for all values of the
variable.
Example: 2(x – 6) = –5x – 12 + 7x.
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
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.
IMPORTANT:
If you multiply or divide both sides by a NEGATIVE
number, you must reverse the direction of inequality
symbol. Example: -6x ≥ 30
-6
-6
x ≤ -5
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
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
Holt Algebra 2
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.
Test x = –4
•
–10 –9
Test x = –2
–8 –7 –6 –5 –4
–3 –2 –1
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.
Holt Algebra 2
–18 ≥ –18 
So –2 is a
solution.
–10 ≥ –5 x
So –1 is not
a solution.
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
Holt Algebra 2
Subtract x from both sides.
Subtract 17 from both sides.
Divide both sides by 3.
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.
Test x = –6
•
–6 –5
Test x = –3
–6 + 8 ≥ 4(–6) + 17 –3 +8 ≥ 4(–3) + 17
2 ≥ –7 
So –6 is a
solution.
Holt Algebra 2
5≥5

So –3 is a
solution.
–4 –3 –2 –1
0
1
2
3
Test x = 0
0 +8 ≥ 4(0) + 17
8 ≥ 17 x
So 0 is not
a solution.