Download One-step equations

Review solving one-step equations with
integers, fractions, and decimals.
One-step equations
Vocabulary
equation
solve
solution
inverse operation
isolate the variable
Addition Property of Equality
Subtraction Property of Equality
An equation uses an equal sign to show that two
expressions are equal. All of these are equations.
3 + 8 = 11
24 = x – 7
100
= 50
2
To solve an equation, find the value of the
variable that makes the equation true. This value
of the variable is called the solution of the
equation.
Additional Example 1: Determining Whether a
Number is a Solution of an Equation
Determine which value of x is a solution of the
equation.
r + 6 = 14
Additional Example 1 Continued
x + 8 = 15; x = 5, 7, or 23
Determine which value of x is a solution of the
equation.
x + 8 = 15; x = 5, 7, or 23
Substitute each value for x in the equation.
Substitute each value for x in the equation.
?
?
x + 8 = 15
?
5 + 8 = 15
?
13=
15 
x + 8 = 15
Substitute 5 for x.
So 5 is not solution.
?
7 + 8 = 15
?
15=
15 
Substitute 7 for x.
So 7 is a solution.
1
Additional Example 1 Continued
Determine which value of x is a solution of the
equation.
x + 8 = 15; x = 5, 7, or 23
Substitute each value for x in the equation.
?
x + 8 = 15
?
23 + 8 = 15
?
31=
15 
Substitute 23 for x.
So 23 is not a solution.
Try This: Example 1
Determine which value of x is a solution of the
equation.
x – 4 = 13; x = 9, 17, or 27
Substitute each value for x in the equation.
?
x – 4 = 13
?
9 – 4 = 13
?
5=
13 
Substitute 9 for x.
So 9 is not a solution.
Try This: Example 1 Continued
Try This: Example 1 Continued
Determine which value of x is a solution of the
equation.
x – 4 = 13; x = 9, 17, or 27
Determine which value of x is a solution of the
equation.
x – 4 = 13; x = 9, 17, or 27
Substitute each value for x in the equation.
Substitute each value for x in the equation.
?
x – 4 = 13
?
17 – 4 = 13
?
Substitute 17 for x.
?
13 = 13 
So 17 is a solution.
Addition and subtraction are inverse
operations, which means they “undo” each
other.
To solve an equation, use inverse operations
to isolate the variable. This means getting
the variable alone on one side of the equal
sign.
x – 4 = 13
?
27 – 4 = 13
Substitute 27 for x.
?
23 = 13 
So 27 is not a solution.
To solve a subtraction equation, like y  15 = 7, you
would use the Addition Property of Equality.
ADDITION PROPERTY OF EQUALITY
Words
You can add the
same number to
both sides of an
equation, and
the statement
will still be true.
Numbers
Algebra
2+3=5
+4 +4
2+7=9
x=y
x+ z = y + z
2
There is a similar property for solving addition
equations, like x + 9 = 11. It is called the
Subtraction Property of Equality.
SUBTRACTION PROPERTY OF EQUALITY
Words
You can subtract
the same number
from both sides of
an equation, and
the statement will
still be true.
Numbers
Algebra
4 + 7 = 11
3 3
4+4= 8
x=y
x z = y  z
Additional Example 2A: Solving Equations Using
Addition and Subtraction Properties
Solve.
A. 10 + n = 18
10 + n = 18
–10
–10
0+n= 8
n= 8
Check
10 + n = 18
?
10 + 8 = 18
Subtract 10 from both sides.
Identity Property of Zero: 0 + n = n.
?
18 = 18 
Additional Example 2B: Solving Equations Using
Addition and Subtraction Properties
Additional Example 2C: Solving Equations Using
Addition and Subtraction Properties
Solve.
Solve.
B. p – 8 = 9
p–8=9
+8 +8
C. 22 = y – 11
22 = y – 11
+ 11
+ 11
Add 8 to both sides.
p + 0 = 17
p = 17
Check
p–8=9
?
17 – 8 = 9
Identity Property of Zero: p + 0 = p.
?
33 = y + 0
Identity Property of Zero: y + 0 = 0.
33 = y
Check
22 = y – 11
?
22 = 33 – 11
?
9 = 9
22 = 22 
Try This: Example 2A
Solve.
A. 15 + n = 29
15 + n = 29
–15
–15
0 + n = 14
n = 14
Check
15 + n = 29
?
10 + 14 = 29
?
29 = 29 
Add 11 to both sides.
Subtract 15 from both sides.
Identity Property of Zero: 0 + n = n.
Try This: Example 2B
Solve.
B. p – 6 = 7
p–6=7
+6 +6
p + 0 = 13
p = 13
Check
p–6=7
?
13 – 6 = 7
Add 6 to both sides.
Identity Property of Zero: p + 0 = p.
?
7 = 7
3
Try This: Example 2C
Solve.
C. 44 = y – 23
44 = y – 23
+ 23
+ 23
Add 23 to both sides.
Learn to solve equations using
multiplication and division.
67 = y + 0
Identity Property of Zero: y + 0 = 0.
67 = y
Check
44 = y – 23
?
44 = 67 – 23
?
44 = 44 
Vocabulary
Division Property of Equality
Multiplication Property of Equality
Additional Example 1: Solving Equations Using
Division
Solve 8x = 32.
8x = 32
8x = 32
8
8
1x = 4
x=4
You can solve a multiplication equation using the
Division Property of Equality.
DIVISION PROPERTY OF EQUALITY
Words
Numbers
Algebra
You can divide both
sides of an equation
by the same
nonzero number,
and the equation
will still be true.
4 • 3 = 12
4 • 3 = 12
2
2
12 = 6
2
x=y
x=y
z z
Try This: Example 1
Solve 9x = 36.
9x = 36
Divide both sides by 8.
1•x=x
Check
8x = 32
? 32
8(4) =
Substitute 4 for x.
?
32 = 32 
9x = 36
9
9
1x = 4
x=4
Divide both sides by 9.
1•x=x
Check
9x = 36
? 36
9(4) =
Substitute 4 for x.
?
36 = 36 
4
Additional Example 2: Solving Equations Using
Multiplication
You can solve a division equation using the
Multiplication Property of Equality.
MULTIPLICATION PROPERTY OF EQUALITY
Words
Numbers
You can multiply
both sides of an
equation by the
same number, and
the statement will
still be true.
2•3=6
4 • 2 • 3 =4 • 6
Algebra
Solve n = 7.
7
n
7•
= 7•7
7
Multiply both sides by 7.
n = 49
x=y
zx =zy
Check
n =7
7
?
49 =
7
7
?
7 = 7
8 • 3 = 24
Substitute 49 for n.
Try This: Example 2
Solve n = 16
4
n
4 • 4 = 4 • 16
Learn to solve equations with integers.
Multiply both sides by 4.
n = 64
Check
n = 16
4
64 =? 16
4
? 16
16 =

Substitute 64 for n.
When you are solving
equations with integers,
your goal is the same as
with whole numbers:
isolate the variable on one
side of the equation.
+
–
+
–
+
–
Additional Example 1A & 1B: Adding and Subtracting
to Solve Equations
0
3 + (–3) = 0
Recall that the sum of a
number and its opposite is 0.
When you add the opposite to
a + (–a) = 0
get 0, you can isolate the
variable.
Solve.
A.
x–3=–6
x–3=–6
x–3+3=–6+3
x=–3
B.
–5 + r = 9
–5 + r = 9
–5 + 5 + r = 9 + 5
r = 14
Add 3 to both sides.
Commutative Property
x–3+3=–3
0
Add 5 to both sides.
5
Additional Example 1C & 1D: Adding and Subtracting
to Solve Equations Continued
Solve.
C. –6 + 8 = n
–6 + 8 = n
2=n
D.
z+6=
z+6=
–6
z=
–3
–3
–6
–9
Solve.
A.
The variable is
already isolated.
B. –2 + g = 5
Add –6 to each side.
Try This: Example 1C & 1D
–1 + 7 = r
–1 + 7 = r
6=r
D.
a + 9 = –9
a + 9 = –9
–9
–9
a = –18
p–7=–9
p–7=–9
p–7+7=–9+7
p = –2
Add integers.
Solve.
C.
Try This: Example 1A & 1B
The variable is
already isolated.
–2 + g = 5
–2 + 2 + g = 5 + 2
g=7
Add 2 to both sides.
Solve.
A.
–5x = 35
–5x = 35
–5
–5
Divide both sides by –5.
x = –7
Add –9 to each side.
Additional Example 2B: Multiplying and Dividing to
Solve Equations Continued
B.
0
Additional Example 2A: Multiplying and Dividing to
Solve Equations Continued
Add integers.
Solve.
Add 7 to both sides.
Commutative Property
p–7+7=–2
Try This: Example 2A
Solve.
z
=5
–4
z
–4 
= –4  5 Multiply both sides by –4.
–4
z = –20
A.
–7x = 42
–7x = 42
–7
–7
Divide both sides by –7.
x = –6
6
Try This: Example 2
Solve.
z
=9
–3
B.
–3 
z
= –3  9 Multiply both sides by –3.
–3
z = –27
Additional Example 1A: Solving Equations with
Decimals.
One-Step Equations
with Rational Numbers
(fractions and decimals)
Additional Examples 1B: Solving Equations with
Decimals
Solve.
B. 8.2p = –32.8
Solve.
A. m + 4.6 = 9
m + 4.6 = 9
Subtract 4.6 from
– 4.6 = – 4.6 both sides.
m = 4.4
8.2p = –32.8
8.2
8.2
Divide both sides by 8.2
p = –4
Remember!
Once you have solved and equation it is a good
idea to check your answer. To check your
answer, substitute your answer for the variable
in the original equation.
Additional Examples 1C: Solving Equations with
Decimals
Solve.
x
C. 1.2 = 15
1.2 •
x
= 1.2 • 15
1.2
x = 18
Multiply both sides by 1.2
Try This: Example 1A & 1B
Solve.
A. m + 9.1 = 3
m + 9.1 = 3
–9.1 = –9.1
Subtract 9.1 from
both sides.
m = –6.1
B. 5.5b = 75.9
75.9
5.5 b
=
5.5
5.5
Divide both sides by 5.5
b = 13.8
7
Try This: Examples 1C
Additional Examples 2A: Solving Equations with
Fractions
Solve.
Solve.
y
C. 4.5 = 90
y
= 4.5 • 90
4.5 •
4.5
A. n + 2 = – 3
7
7
Multiply both sides by 4.5
n–
2 2
3 2
+ =– –
7 7
7 7
y = 405
n=–
Additional Examples 2B: Solving Equations with
Fractions
Solve.
Solve.
C. 5 x = 5
Add
1
to both sides.
6
Find a common denominator; 6.
Simplify.
5
6
5
63
•
x= 8 •
5
6
5
4
3
4
Try This: Example 2A
9
9
1 1
5 1
n– + =– –
9 9
9 9
2
n=–
3
1
Subtract 9 from both sides.
Simplify –
6
.
9
5
7
8
6
x =
Solve.
A. n + 1 = – 5
2
from both sides.
7
Additional Examples 2C: Solving Equations with
Fractions
B. y – 1 = 2
6
3
1
1
2 1
+y– = +
6
6
3 6
4 1
y= +
6 6
y= 5
6
Subtract
Multiply both sides by
6
.
5
Simplify.
Try This: Example 2B
Solve.
1
3
B. y – 2 =
4
1 y 1
3 1
+ – = +
2
2
4 2
3 2
+
4 4
1
y=1
4
y=
Add
1
to both sides.
2
Find a common denominator; 4.
Simplify.
8
Try This: Examples 2C
Solve.
6
3
C. 8 x = 19
6
3x
=
19
8
2
6
8
3
8
• x =
•
19 3 1
8
3
x =
16
19
Multiply both sides by
8
.
3
Simplify.
9