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2.1 THE ADDITION PROPERTY OF EQUALITY
We’re going to be solving equations in “one variable.” The answer(s)
we get are called
 “Solutions” (or “roots”) of the equations.
 We say that the answer(s) “satisfy” the equation (they are the
values that make the equation true).
A linear equation in one variable (x) is an equation that can be written
in the form 
ax + b = c
Where a, b and c are real numbers, and a  0 (“a” is not equal to zero).
Two things to notice about this linear equation:
 the exponent on the x is always 1
 there is only going to be 1 variable
Review the following examples. Compare them to the form ax + b = c.
4x + 2 = 10
what does a=______, b=______, c=_______
-3x =19
what does a=______, b=______, c=_______
x = 13
what does a=______, b=______, c=_______
FYI – you need to understand a few things about exponents before we
go on. We’re going to start by looking at a pattern:
Keep this chart handy.
25  2  2  2  2  2  32
2 4  2  2  2  2  16
23  2  2  2  8
22  2  2  4
21  2
20  1
1
2 1 
2
1 1 1
22   
2 2 4
1 1 1 1
2 3    
2 2 2 8
1 1 1 1 1
24     
2 2 2 2 16
More to remember:
1. Whenever you see a variable and no exponent is shown, it’s
understood that there is an exponent of one
x  x1
x1  x
1. Part of the reason these equations are called “linear” is
because, if we were to graph them, we would get a straight line.
When solving linear equations, we want to manipulate the equation so
we end up with the last line saying “x = some number.” That last part
(“some number”) is the solution. We have to follow some rules in order
to manipulate the equation correctly.
ADDITION PROPERTY OF EQUALITY
Says you can take any equation, add the same real number to
BOTH sides and you end up with an equivalent equation.
In general:
a=b
a+c=b+c
These two equations would be equivalent to each other.
A specific example:
x=9
x+5=9+5
The equation remains true (balanced) as long as you do the same to
BOTH sides.
Now let’s solve an equation using the rules above—
x – 11 = 8
Remember, we ultimately want to find x.
So, first we have to get rid of the -11, by
adding 11 to both sides. The reason we pick
11 is because that’s the only number, when
added to -11 that will equal 0. Then we’ll
just have x + 0 which equals x.
x – 11 + 11 = 8 + 11
x + 0 = 19
x = 19
To check the answer, we take the final solution and substitute it
back in to the original equation. If we end up with a true
statement, then our solution is correct.
Check:
x – 11 = 8
19 – 11 = 8
8 = 8 true, so the solution is correct.
When you provide me written work, the following format is what I
expect to see!
Terminology – Sometimes we/your book may refer to an answer as the
“solution set.” It’s just another type of mathematical terminology used
when referring to an answer. If you were asked to provide the solution
set for the following equation, you would provide your answer using set
notation (braces).
Solve the following equation:
x–4=3
Check: x – 4 = 3
x–4+4=3+4
7–4=3
x+0=7
3=3
x=7
true
Caution!
7 is the solution
{7} is the solution set
We can use the rule for ADDITION PROPERTY OF EQUALITY when
we need to subtract a number from both sides of an equation.
Remember, subtraction is the same as addition, you’re just adding the
opposite of the number (you can change any subtraction problem to an
addition problem).
Example:
y + 1.5 = 4.5
y + 1.5 – 1.5 = 4.5 – 1.5
y + 0 = 3.0
y = 3.0 or
y=3
Example:
1
2
 z
2
3
1 2
2 2
  z 
2 3
3 3
3 4
  z0
6 6
7
z
6
7
z
6
Check: y + 1.5 = 4.5
3.0 + 1.5 = 4.5
4.5 = 4.5
true, so our solution is correct
Check:
1
2
1
2
1
2
1
2
1
2
2
3
7 2

6 3
7 4

6 6
3
6
1
2
 z




True.
So 7 is correct.
6
Now, we’ll see a variable on both sides of the equation.
7 x  3  6( x  1)  9
You first must apply the distributive property.
Then we can proceed to solve.
7 x  3  6x  6  9
7 x  3  6x  3
Get the x’s on one side; then combine like
7 x  3  6x  6x  3  6x
terms.
x3 03
Get rid of the 3 on the left side of the equation.
x33
x 33  33
x0
Now, we check the answer.
Remember, you must substitute back in to the original equation.
7 x  3  6( x  1)  9
7  0  3  6(0  1)  9
0  3  6(1)  9
3  6  9
33
True. So, our solution of “0” is correct.
Homework: 30, 38, 52, 62, 68 Updated 012912.