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Transcript
MTH 10905
Algebra
THE ADDITION PROPERTY OF
EQUALITY
CHAPTER 2 SECTION 2
Identify Linear Equations
 Equations are statements that show two algebraic expressions are
equal

2x + 1 = x – 5 is an example of an equation
 Linear Equation in one variable is an equation that can be written
in the form
ax + b = c
a, b and c are real numbers and c ≠ 0
Linear Equations in other words is an equation which has 1
variable that is multiplied by a number and some constant. It can
also have the variable on both sides of the equation. For example:
x + 4 = 2x – 6
Identify Linear Equations
 Examples of Linear Equations:
Exp:
Exp:
Exp:
Exp:
Exp:
4x – 3 = 5
x + 1 = -6
x + 5 = 5x + 2
x + 9 = 52
27 = x + 16
Identify Linear Equations
 Solve an Equations is to find the number that when
substituted for the variable makes the equation true.
 Checked is when we substitute the answer that is
believed to be the answer into the original equation.
 We use a
? when we are checking to see if this is a
true statement.
 When solving or checking keep your equal signs in line
to help you follow your work.
Linear Equations
 Exp:
Consider the equation 6x – 5 = 25 is 3 a solution
6 x  5 ? 25
6(3)  5 ? 25
18  5 ? 25
13  25
No, 3 is not a solution
Linear Equations
 Exp:
Determine whether 35 is a solution to the equation
4x – 3(x – 5) = 50
4x  3( x  5)  50
4(35)  3(35  5)  50
4(35)  3(30)  50
140  90  50
50  50
Yes, 35 is a solution
Linear Equations
 Exp:
Determine whether
4(x+ 3) = 7 + x
LCD = 3
12 3 36

1 3 3
7 3 21

3 3 3
Yes,

5
3
is a solution

5
3
is a solution to the equation
4 x  (4)(3)  7  x
 5
 5
4     12  7    
 3
 3
-
20 12 7 5
  
3 1 1 3
20 36 21 5

 
3
3
3 3
20  36 21  5

3
3
16 16

3
3
Identify Equivalent Equations
 Solve an Equation to solve an equation you have to get the variable
alone on one side of the equal sign or isolate the variable.
 Equation must be in the form of x = some number
 To ensure that an equation remains equal or balanced we have to do the
same thing to both sides of the equation.
 If I have a balanced scale and I add 3 lbs to one side what do I have to
do to keep it balanced?
 Adding, subtracting, multiplying or dividing a number to the left side
means to add, subtract, multiply or divide the same number to the right
side.
Addition Property of Equality
 Equivalent equations are two or more equations with the same solution.
Exp: 3x + 3 = 6 and 3x = 3 equivalent because the solution to both
is 1
 To solve an equation we can use the addition property of equality or
the multiplication property of equality
 The addition property says if a = b then a + c = b + c for any real
number a, b, and c
 The addition property of equality is used to make an equivalent
equation and when used correctly t can be used to solve an equation
Addition Property of Equality
 The multiplication property says if a = b then a · c = b · c for any
real number a, b, and c
 Addition property is used to solve equation in the form of
x+a=b
 When we add or subtract to the left side we must add or subtract to
the right side to keep the equation equal. This eliminates the number
on the same side of the equal sign as the variable.
 Subtraction is defined in terms of addition, therefore the addition
property allows us to also subtract on both sides of an equation
Addition Property of Equality
 Exp:
Exp:
x + 3 = 10
x + 3 – 3 = 10 – 3
x=7
x–4=5
x–4+4=5+4
x=9
CHECK:
x + 3 = 10
7 + 3 = 10
10 = 10
CHECK:
x–4=5
9–4=5
5=5
Addition Property of Equality
 Exp:
Exp:
a – 8 = -12
a + 8 – 8 = -12 + 8
x = -4
t+4=7
t–4+4=7–4
t=3
CHECK:
CHECK:
a – 8 = -12
-4 – 8 = -12
-12 = -12
t+4=7
3+4=7
7=7
Addition Property of Equality
 Exp:
-5 = b + 10
CHECK:
-5 – 10 = b + 10 – 10
-15 = b
-5 = b + 10
-5 = -15 + 10
-5 = -5
 Remember that our goal is to isolate the variable on one side of
the equation. To do this we add or subtract the number on the
same side as the variable to both sides of the equation.
 Exp:
-10 = x – 3
-10 + 3 = x
-7 = x
CHECK:
-10 = x – 3
-10 = -7 – 3
-10 = -10
Addition Property of Equality
 Exp:
-8.75 = r + 13.25
-8.75 – 13.25 = r + 13.25 – 13.25
-22.00 = r
CHECK:
-8.75 = r + 13.25
-8.75 = -22.00 + 13.25
-8.75 = -8.75
HOMEWORK 2.2
 Page 111 – 112
#25, 29, 33, 53, 57, 67