Solving Equations
The equations are equivalent If they have the
same solution(s)
Example:
Determine whether 4x = 12 and 10x = 30 are
equivalent equations
Determine whether 3x = 4x and 3/x = 4/x are
equivalent equations
ADDITION AND MULTIPLICATION
PROPERTIES OF EQUALITY
 Let a, b, and c represent algebraic expressions
 Addition property of equality:
If a = b,
then a + c = b + c
 Multiplication property of equality:
If a = b,
then a(c) = b (c)
APPLYING THE ADDITION
PROPERTIE OF EQUALITY
In each equation, the goal is to isolate the variable on
one side of the equation. To accomplish this, we use
the fact that the sum of a number and its opposite is
zero and the difference of a number and itself is zero.
p – 4 = 11
To isolate p, add 4 to
both sides (-4 +4 = 0).
p – 4 +4 = 4 +4
p- + 0 = 15
p = 15
Simplify
CHECK
Let’s try:
Solve : y  4.7  13.9
y  4.7  13.9
y  4.7  4.7  13.9  4.7
Using the addition principle, adding 4.7 to both sides
y  0  13.9  4.7
The solution of this equation is 18.6
y  18.6
Check :
18.6  4.7  13.9
13.9  13.9
Substituting 18.6 for y
TRUE
Applying the Multiplication
Properties of Equality
12x = 60
To obtain a coefficient
of 1 for the x-term,
divide both sides by 12
12x = 60
12 12
Simplify
x=5
Check!
Tip: Recall that
the product of a
number and its
reciprocal is 1.
For example:
1
(12)  1
12
Example:
2
1
 q
9
3
Tip: When applying the multiplication or division
properties of equality to obtain a coefficient of 1 for
the variable term, we will generally use the following
convention:
1. If the coefficient of the variable term is
expressed as a fraction, we usually multiply both
sides by its reciprocal.
2. If the coefficient of the variable term is an
integer or decimal, we divide both sides by the
coefficient itself.
9
2
1 9
( )( q )  ( )
2
9
3 2
3
q
2
To obtain a coefficient of 1 for the q- 2
term, multiply by the reciprocal of  9
which is  9
2
Simplify. The product of a number
and its reciprocal is 1.
CHECK!
Let’s try:
Solve : 8t  72
8t  72
Using the multiplication
1
1
8( )t  72  
both sides
8
8
72
t
8
t 9
The solution of this equation is 9
Check :
8(9)  72
72  72
principle, multiply by 1/8 to
Substituting 9 for t
TRUE
Steps to solve a linear Equation in One Variable
 Simplify both sides of the equation.




Clear parentheses
Combine like terms
Use the addition or subtraction property of equality to
collect the variable terms on one side of the equation.
Use the addition or subtraction property of equality to
collect the constant terms on the other side of the
equation.
Use the multiplication or division property of equality to
make the coefficient of the fvariable term equal to 1.
Check your answer.
We will want to find those values of x that make the equation
true by isolating the x (this means get the x all by itself on
one side of the equal sign)
2x  4  52 x  1
2 x  8  10 x  5
+8
+8
2 x  10x  3
+ 10x
+ 10x
12 x  3
12
12
+3
Since the x is in more than
one place and inside of
parenthesis the first thing
we’ll do is get rid of
parenthesis by distributing.
Now let’s get all constants
(terms without x’s) on the right
side. We’ll do this by adding 8
to both sides.
We are ready to get all x terms on the
left side by adding 10x to both sides.
1
x
4
Now get the x by itself by getting
rid of the 12. 12x means 12 times
x so we get rid of it by dividing
both sides by 12.
Let’s check this answer by substituting it into the original
equation to see if we get a true statement.
2

?
1
x4  4 
5

1
2 x4  1
 
?
1
1 
 8  5  1
2
2 
1
5
?
8   5
2
2
1 16 ? 5 10
  
2 2
2 2
Distribute and multiply
Distribute
Get a common denominator
15
15
 
2
2
It checks!
Let’s try:
3( x  4)  7 x
3 x  12  7 x
3 x  3 x  12  7 x  3 x
Using the addition principle, adding 4.7 to both sides
12  4 x
1
1
( )12  ( )4 x
4
4
3 x
TRUE
Using the multiplication principle, multiply by 1/4 to
both sides
Conditional Equations, Identities,
and Contradictions
 Conditional Equations
An equation that is true for some values of the variable
but false for other values.
 Contradictions
Equation with no solution
 Identities
An equation that has all real numbers as its solutions.
Classify the equation
3  8x  5  7 x
Using the addition principle
3  8x  7 x  5  7 x  7 x
Simplifying
3 x  5
Using the addition principle
3  3  x  5  3
Simplifying
x  2
Divide both sides b -1.
2
x  , or 2
1
There is one solution, -2. For other choices of x, the equation is
false. This equation is conditional since it can be true or false,
depending on the replacement for x.
Classify the equation
3 x  5  3( x  2)  4
Using the distributive law
3x  5  3x  6  4
3x  5  3x  2
Combining like terms.
3 x  3 x  5  3 x  3 x  2
Using the addition principle, adding -3x to both sides
5  2
The equation is false regardless of what x is replaced with, so all
real numbers are solutions. There is no solution. This equation
is a contradiction.
Classify the equation
3 x  5  3( x  2)  4
Using the distributive law
3x  5  3x  6  4
3x  5  3x  2
Combining like terms.
3 x  3 x  5  3 x  3 x  2
Using the addition principle, adding -3x to both sides
5  2
The equation is false regardless of what x is replaced with, so all
real numbers are solutions. There is no solution. This equation
is a contradiction.
Let’s try:
7 x  2  3x  4 x
2  9 x  3(4 x  1)  1