Solving One-Variable Equations

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Algebra 1 Notes SOL A.4 Equations
Mr. Hannam
Name: __________________________________________ Date: _______________ Block: _______
Equations
 An equation is an open sentence where two expressions are set _____________.

Equations may have one or more unknowns (_______________) which we may try to solve.

Solving an equation means finding the value(s) of variable(s) that make the equation
____________.

Equations that have the same solution(s) are called ____________________.

We use ______________________ operations to solve equations.

We justify the steps in solving equations by using field properties.
Solving Equations

“Isolate” the variable, justifying steps using field properties (properties of equality):
1) Put variables on one side of the = and numbers on the other by isolating x:

perform inverse operations (add, subtract, multiply, or divide)
2) Whatever you do to one side of the equation, you do to the other
3) Verify solutions
S

substitute solution in original equation (DO NOT SKIP THIS STEP!!!!!!)
Field Properties of Equality:
Property of Equality
Reflexive Property of Equality
Algebra (for real numbers a, b, c)
a=a
Symmetric Property of Equality
If a = b, then b = a
Transitive Property of Equality
If a = b and b = c, then a = c
Substitution Property of Equality
If a = b, then a can be substituted for b
Addition Property of Equality
If a = b, then a + c = b + c
Subtraction Property of Equality
If a = b, then a – c = b - c
Multiplication Property of
Equality
If a = b, then ac = bc
Division Property of Equality
If a = b and c  0, then
a b

c c
Example
Algebra 1 Notes SOL A.4 Equations

Solve the equations:
a) x – 4 = 6
x–4+4=6+4
x = 10
VERIFY: Is 10 a solution?
(10) – 4 = 6
6=6


Mr. Hannam Page 2
b) x + 3 = -5
x + 3 – 3 = -5 – 3
x = -8
VERIFY: Is -8 a solution?
(-8) + 3 = -5
-5 = -5
c) 4x = 16
4x ÷ 4 = 16 ÷ 4
x=4
VERIFY: Is 4 a solution?
4(4) = 16
16 = 16
Solve the equations, justifying steps:
a) Solve x + 4 = 10
1) x + 4 = 10
2) x + 4 – 4 = 10 – 4
3) x = 6
b) Solve
Given
Subtraction Property
of Equality
Simplify
2
x=4
3
2
x=4
Given
3
3 2
3
 x   4 Mult. Property of Equality
2)
2 3
2
3) x = 6
Simplify
1)
Two-Step Equations

SAME! Put variables on one side of the = and numbers on the other:
How to Solve Two-Step Equations
1) Clear parentheses (distribute) and combine like terms if necessary
2) Do add/subtract steps first
3) Do multiply/divide steps
Examples:
x
 5  11
2
x
 5  5  11  5 ________________
2
x
 16
________________
2
x
2   2  16
________________
2
x = 32
_________________
VERIFY:
a) 5x + 9 = 24
5x + 9 – 9 = 24 – 9 __________________
5x = 15
__________________
5x 15

__________________
5
5
x=3
__________________
VERIFY:
b)
c) 3x + 2x = 15
d) 4(x - 6) = 32
VERIFY:
VERIFY:
Algebra 1 Notes SOL A.4 Equations
Mr. Hannam Page 3
Practice - solve the equations, justifying steps with field properties:
a) x + 9 = 25
b) -4x = -20
a
46
3
e)
f) -1 =
z
3
37
3
x  3
5
d) 2x + 3 = -9
g) 7x – 4x = 21
h) 3(x + 2) = 9
c)
Multistep Equations and Equations with Variables on Both Sides
How to Solve Multi-Step Equations and Equations with Variables on Both Sides
1) Clear parentheses (distribute) and combine like terms if necessary
2) Add/subtract variable terms so that variable is on one side (NEW STEP!)
3) Do add/subtract steps first
4) Do multiply/divide steps
Example: Solve 7 – 3x = 4x – 7
1) 7 – 3x = 4x – 7
Given
2) 7 – 3x - 4x = 4x – 7 - 4x
________________________________________________
3) 7 – 7x = - 7
________________________________________________
4) 7 -7x - 7 = -7 - 7
________________________________________________
5) -7x = -14
________________________________________________
6)
________________________________________________
- 7x - 14

7
-7
7) x = 2
________________________________________________
Examples: Solve the equation, justifying steps…
1
a) 9x – 5 = (16x + 60)
b) 3(x + 12) – x = 4(2 – x) – 3x + 2x
4
9x – 5 = 4x + 15 ______________________
5x – 5 = 15
______________________
5x = 20
______________________
x=4
_____________________
Algebra 1 Notes SOL A.4 Equations
Mr. Hannam Page 4
Special cases:
Identity case (solutions are all real numbers):
No solution case:
2x + 6 = 2(x + 3)
3x = 3(x + 4)
2x + 6 = 2x + 6
Distribute
3x = 3x + 12 Distribute
0=0
Subtraction Prop. of =
0 = 12
Subtraction Prop. of =
Where did the variable go??
Where did the variable go??
When we get a TRUE statement at the end
when the variable “disappears,” EVERY x is
a solution (the two sides of the equation are
identical!).
When we get a FALSE statement at the end
when the variable “disappears,” there are NO
SOLUTIONS. There is no value of x that
makes the equation true.
ALL REAL NUMBERS are solutions.
Other Equations:
Solve: a)
x 3

4 2
Cross Product Property:

b)
If
x
2

x 1 3
a c
 then ad = bc
b d
REMEMBER TO GROUP NUMERATORS AND DENOMINATORS (use parentheses)
You Try: Solve the equation if possible; if not, write “all real numbers” or “no solution”…
a) 8x + 5 = 6x + 1
b) x + 1 = 3x - 1
c) 9x = 6(x + 4)
d) 7(x + 7) = 5x + 59
e) 2 – 15x = 5(-3x + 2)
f) 12y + 6 = 6(2y + 1)
h) 40 + 14x = 2(-4x – 13)
i) 2(3x + 2) =
g) 5(x + 2) =
j)
x 9

2 3
3
(5 + 10x)
5
k)
4
8

x -8 2
l)
2
x

5 21 - x
1
(12x + 8)
2
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