7.2-3 Solving Linear Equations
A linear equation in one variable is
an equation in which the same letter
is used in all variable terms and the
exponent of the variable is 1.
The solution to an equation in
one variable is the number that
can be substituted in place of the
variable and makes the equation
true.
For example 5 is a solution to
the equation 2x + 3 = 13 because
2(5) + 3 = 13 is true.
Equivalent equations are
equations that have the same
solutions. For example
2x + 3 = 13
and x = 5 are equivalent
equations because each has the
solution of 5.
Basic Principle of Equality
• To preserve equality, if an operation is
performed on one side of an equation, the
same operation must be performed on the
other side.
There are 2 principles (axioms) we
will use to solve linear equations in
one variable. The first is the
addition principle of equality. This
principle allows us to add (or
subtract) the same value to both
sides of an equation to obtain an
equivalent equation.
To solve an equation using the
addition axiom:
• Locate the variable in the equation.
• Identify the constant that is associated with
the variable by addition or subtraction.
• Add the opposite of the constant to both
sides of the equation.
Solve each equation.
y

3

11
y

7
.
2

2
.
3
x 896x1y35
Often we need to combine like
terms on one or both sides of the
equation before solving.
For example:
7  3x  4  5x  9x  6
combining like terms gives
x  3  6
Whenever variable terms appear on
both sides of an equation we use the
addition principle to move all variable
terms to the same side, then solve.
For example to solve
 5x  8  6 x
add 6x to both sides to get
x 8
When solving these type equations, it
makes no difference the side from
which you remove the variable term to
start. The goal is to get all variable
terms on one side and all constants on
the other.
Solve each equation.
5x  7  4 x
7 x  6x  5
3c  8  2c  5
Whenever quantities appear in
parentheses on either side of the
equation they must be removed first.
3x  8  2 x  1
3x  24  2x  1
x  24  1
Solve each equation.
7
283u 
4x x419
63u3xx 
2
17
The Multiplication Principle of
Equality
For each problem so far the
coefficient for the variable ended
up being one. We use the
multiplication principle (axiom)
to solve equations where the
coefficient of the variable is not
one.
The multiplication principle allows
us to multiply (or divide) each side of
an equation by the same nonzero
quantity to obtain an equivalent
equation.
The goal is to get
+1 times the variable = a number.
To solve an equation where the
coefficient of the variable is not
one you need to multiply both
sides of the equation by the
reciprocal of the coefficient. An
alternate way is to divide both
sides by the coefficient of the
variable.
Solve each equation.
3
2
3
4x
5
xqx
25
83
24
2
.
811
m
16
.
8
42
6
x
x

0
y


4

15
x

16
 72


18
w
4
5
7
9
Sometimes it is necessary to
combine like terms before
solving the equation.
8x  3x  45 combine like terms
5x  45 divide both sides by 5
x 9
Solve each equation.
9
ws26ws 
ws48
52
 576
6
When more than one operation is
indicated on the variable, undo
addition or subtraction first, then undo
multiplication or division next.
Solve: 4x – 2 = 18
• Since the variable has
been multiplied by 4
and subtracted by 2,
undo by adding 2 and
dividing by 4.
4 x  2  2  18  2
4 x 20

4
4
x5
4  5   2  18
18  18
Solve: 118 – 22m = 30
• Think of 118 – 22m as
118 + ( - 22m)
118  22m  118  30  118
22m  88
22m 88

22
22
m4
118  22  4   30
118  88  30
30  30
Solve: 5x – 4 = 8x – 13
5 x  4  5 x  8 x  13  5 x
4  3 x  13
4  13  3 x  13  13
9  3x
9 3x

3 3
3 x
5 x  4  8 x  8 x  13  8 x
3 x  4  13
3 x  4  4  13  4
3 x  9
3 x 9

3 3
x3
Summary of steps for solving an
equation:
• Remove parentheses.
• Combine like terms on each side of the
equation.
• Sort terms to collect the variable terms on
one side and constants on the other.
• Solve for the variable by multiplying by the
reciprocal of the coefficient or dividing by
the coefficient of the variable.
Solve the following equation.
Step 1
Step 2
Step 3
5(h – 4) + 2
5h – 20 + 2
5h – 18
5h – 18 + 18
5h
5h – 3h
2h
2h
2
h
=
=
=
=
=
=
=
3h
3h
3h
3h
3h
3h
14
= 14
2
= 7
–
–
–
–
+
+
4
4
4
4 + 18
14
14 – 3h
Distribute.
Combine terms.
Add 18.
Combine terms.
Subtract 3h.
Combine terms.
Divide by 2.
Step 4
Check by substituting 7 for h in the original equation.
5 ( h – 4 ) + 2 = 3h – 4
5 ( 7 – 4 ) + 2 = 3(7) – 4
?
Let h = 7.
5 (3) + 2 = 3(7) – 4
?
Subtract.
15 + 2 = 21 – 4
?
Multiply.
17 = 17
The solution to the equation is 7.
True
Solving an Equation That Has Infinitely
Many Solutions
4 ( 2n + 6 ) = 2 ( 3n + 12 ) + 2n
8n + 24 = 6n + 24 + 2n
Distribute.
8n + 24 = 8n + 24
Combine terms.
8n + 24 – 24 = 8n + 24 – 24
8n = 8n
8n – 8n = 8n – 8n
0 = 0
Subtract 24.
Combine terms.
Subtract 8n.
True
An equation with both sides exactly the same, like 0 = 0, is called an
identity. An identity is true for all replacements of the variables. We
indicate this by writing all real numbers.
Solving an Equation That Has No Solution
6x – 1 ( 4 – 3x )
6x – 4 + 3x
9x – 4
9x – 4 – 9x
–4
=
=
=
=
=
8 +
8 +
–19
–19
–19
3 ( 3x – 9 )
9x – 27
+ 9x
+ 9x – 9x
Distribute.
Combine terms.
Subtract 9x.
False
Again, the variable has disappeared, but this time a false statement
(– 4 = – 19) results. Whenever this happens in solving an
equation, it is a signal that the equation has no solution and we
write no solution.