Solving equations using addition and subtraction

Solving Equations by
2-1 Adding or Subtracting

Objective
Solve one-step equations in one
variable by using addition or
subtraction.

Holt Algebra 1


Vocabulary
equation
solution of an equation

Holt Algebra 1


An equation is a mathematical statement that
two expressions are equal.

A solution of an equation is a value of the
variable that makes the equation true.

To find solutions, isolate the variable. A
variable is isolated when it appears by itself
on one side of an equation, and not at all on
the other side.

Holt Algebra 1


Isolate a variable by using inverse operations which
"undo" operations on the variable.
An equation is like a balanced scale. To keep the
balance, perform the same operation on both sides.

Inverse Operations
Operation Inverse Operation
Addition Subtraction

Subtraction Addition

Holt Algebra 1

Example 1A: Solving Equations by Using Addition

Solve the equation. Check your answer.
y – 8 = 24
+8 +8 Since 8 is subtracted from y, add 8 to
y = 32 both sides to undo the subtraction.

Check y – 8 = 24 To check your solution,
32 – 8 24 substitute 32 for y in the
24 24  original equation.

Holt Algebra 1

Example 1B: Solving Equations by Using Addition

5 =z– 7 7
16 16 Since is subtracted from z, add 7 to
16 16
+ 7 + 7 both sides to undo the subtraction.
16 16
3=z
4
5
=z– 7
Check 16 16
7 To check your solution,
5 3
16
–
4 16 substitute 3 for z in the
4
5 5 original equation.
16 16 
Holt Algebra 1

Check It Out! Example 1a

n – 3.2 = 5.6
+ 3.2 + 3.2 Since 3.2 is subtracted from n, add 3.2
n = 8.8 to both sides to undo the subtraction.

Check n – 3.2 = 5.6 To check your solution,
8.8 – 3.2 5.6 substitute 8.8 for n in the
5.6 5.6  original equation.

Holt Algebra 1

Check It Out! Example 1b

–6 = k – 6
+6 + 6 Since 6 is subtracted from k, add 6 to
0=k both sides to undo the subtraction.

Check –6 = k – 6 To check your solution,
–6 0 – 6 substitute 0 for k in the
–6 –6  original equation.

Link 1
Holt Algebra 1

Example 2A: Solving Equations by Using Subtraction

m + 17 = 33
– 17 –17 Since 17 is added to m, subtract 17
m = 16 from both sides to undo the addition.

Check m + 17 = 33 To check your solution,
16 + 17 33 substitute 16 for m in the
33 33  original equation.

Holt Algebra 1

Example 2B: Solving Equations by Using Subtraction

4.2 = t + 1.8
–1.8 – 1.8 Since 1.8 is added to t, subtract 1.8
from both sides to undo the addition.
2.4 = t

Check 4.2 = t + 1.8 To check your solution,
4.2 2.4 + 1.8 substitute 2.4 for t in the
4.2 4.2  original equation.

Holt Algebra 1


d+ 1 =1
2
1 1
– 1 – 1
2
Since 2 is added to d, subtract 2
2
from both sides to undo the addition.
d= 1
2

Check d+ 1 =1 To check your solution,
2 1
1 + 1 substitute 2 for d in the
1
2 2 original equation.
1 1
Holt Algebra 1


–5 = k + 5
–5 – 5 Since 5 is added to k, subtract 5 from
–10 = k both sides to undo the subtraction.

Check –5 = k + 5 To check your solution,
–5 –10 + 5 substitute –10 for k in the
–5 –5  original equation.

Link 2
Holt Algebra 1


Remember that subtracting is the same
as adding the opposite. When solving
equations, you will sometimes find it
easier to add an opposite to both sides
instead of subtracting.

Holt Algebra 1

Example 3: Solving Equations by Adding the Opposite
2
Solve – 5 + p = – .
11 11
+ 5 + 5
5
Since – is added to p, add 5
11 11 11 11
to both sides.
3
p=
11

Check – 5 +p=– 2
11 11 To check your solution,
5 3 – 2 substitute 3 for p in the
– + 11
11 11 11 original equation.
2 2
– –
11 11 
Holt Algebra 1


Solve –2.3 + m = 7. Check your answer.
–2.3 + m = 7
+2.3 + 2.3 Since –2.3 is added to m,
m = 9.3 add 2.3 to both sides.

Check –2.3 + m = 7 To check your
–2.3 + 9.3 7 solution, substitute
9.3 for m in the
7 7
original equation.

Holt Algebra 1

5
Solve – 3 + z = . Check your answer.
4 4
3 +z= 5
–
4 4
3 + 3 Since – 3 is added to z, add 3
+ 4 4
4 4 to both sides.
z=2

Check – 3 +z= 5 To check your solution,
4 4 substitute 2 for z in the
– 3 +2 5 original equation.
4 4
5 5 
4 4
Holt Algebra 1

Check It Out! Example 3c

Solve –11 + x = 33. Check your answer.
–11 + x = 33
+11 +11 Since –11 is added to x,
x = 44 add 11 to both sides.

Check –11 + x = 33 To check your
–11 + 44 33 solution, substitute 44
for x in the original
33 33 
equation.
Link 3
Holt Algebra 1


Properties of Equality
Addition Property of Equality
You can add the same number to both
WORDS
sides of an equation, and the statement
will still be true.
3=3
NUMBERS 3+2=3+2
5=5
a=b
ALGEBRA
a+c=b+c
Holt Algebra 1


Properties of Equality
Subtraction Property of Equality
You can subtract the same number from
WORDS
both sides of an equation, and the
statement will still be true.
7=7
NUMBERS 7–5=7–5
2=2
a=b
ALGEBRA
a–c=b–c
Holt Algebra 1

Solving equations using addition and subtraction

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