aics9e_ppt_2 _1.ppt

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Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 2-1
Equations and
Inequalities
Chapter 2

2
2.1 – Solving Linear Equations
2.2 – Problem Solving and Using Formulas
2.3 – Applications of Algebra
2.4 – Additional Application Problems
2.5 – Solving Linear Inequalities
2.6 – Solving Equations and Inequalities
Containing Absolute Values
Chapter Sections

3
§ 2.1
Solving Linear
Equations

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Properties of Equality
Properties of Equality
For all real numbers a, b, ,and c:
1. a = a Reflexive property
2. If a = b, then b = a Symmetric property
3. If a = b, and b = c, then a = c Transitive property

5
Like terms are terms that have the same
variables with the same exponents.
Like Terms
-3x, 8x, - x
6w2, -12w2, w2
3
1
Unlike Terms
20x, x2, x3
6xy, 2xyz, w2
Combine Like Terms

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Combining Like Terms
1. Determine which terms are like terms.
2. Add or subtract the coefficients of the like
terms.
3. Multiply the number found in step 2 by the
common variable(s).
Example: 5a + 7a = 12a

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Distributive Property
For any real numbers a, b, and c,
a(b + c) = ab + bc
Example: 3(x + 5) = 3x + 15
(This is not equal to 18x! These are
not like terms.)

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Simplifying an Expression
1. Use the distributive property to remove
any parentheses.
2. Combine like terms.
Example:
Simplify 3(x + y) + 2y
= 3x + 3y + 2y (Distributive Property)
= 3x + 5y (Combine Like Terms)
(Remember that 3x + 5y cannot be combined because
they are not like terms.)

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Solve Linear Equations
A linear equation in one variable is an
equation that can be written in the
form ax + b = c where a, b, and c are real
numbers and a  0.
The solution to an equation is the number
that when substituted for the variable
makes the equation a true statement.

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Example:
Solve the equation x – 4 = -10.
x – 4 = -10
x – 4 + 4 = -10 + 4 (Add 4 to both sides.)
x = -6
Check: (-6) – 4 = -10
Addition Property of Equality
If a = b, then a + c = b + c for any
real numbers a, b, and c.

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Multiplication Property of Equality
If a = b, then a · c = b · c for any real
numbers a, b, and c.
Example: Solve the equation 12y = 15.
·12y = 15 · (Multiply both sides by )
12
1
12
1
12
1
y
  
1 1
12 15
12 12
4
5
1
1
y  5
4

12
x
   
1 3
( 4)( ) ( 4)
4 4
(Multiply both sides by -4)
Example: Solve the equation .
x
 
1 3
4 4
x
   
1 3
( 4)( ) ( 4)
4 4
(Simplify)
x = -3
4
3
(-3)
4
1 

Check:
Multiplication Property of Equality

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1. Clear fractions. If the equation contains fractions,
eliminate the fractions by multiplying both sides of the
equation by the least common denominator.
2. Simplify each side separately. Simplify each side of
the equation as much as possible. Use the distributive
property to clear parentheses and combine like terms
as needed.
3. Isolate the variable term on one side. Use the
addition property to get all terms with the variable on
one side of the equation and all constant terms on the
other side. It may be necessary to use the addition
property a number of times to accomplish this.

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4. Solve for the variable. Use the multiplication
property to get the variable (with a coefficient
of 1) on one side.
5. Check. Check by substituting the value
obtained in step 4 back into the original
equation.

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Solving Equations
Example: Solve the equation 2x + 9 = 14.
Don’t forget to check!
2
5
2
5
2
2
5
2
9
14
9
9
2
14
9
2









x
x
x
x
x

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Solve Equations Containing Fractions
Example: Solve the equation 9
3
2
5 


a
27
2
15
27
3
2
3
)
5
(
3
)
9
(
3
3
2
5
3
9
3
2
5



















 



a
a
a
a
21
2
42
2
2
42
2
15
27
2
15
15














a
a
a
a
The least common denominator is 3.

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Identify Conditional Equations, Contradictions,
and Identities
Conditional Equations: Equations that true for only specific values
of the variable.
Contradictions: Equations that are never true and have no solution.
Identities: Equations that are always true and have an infinite
number of solutions.

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Solving Equations with Decimals
Example: Determine whether the equation 5(a - 3) – 3(a –
6) = 2(a + 1) + 1 is a conditional equation, a contradiction,
or an identity.
3
2
3
2
1
2
2
18
3
15
5
1
)
1
(
2
)
6
(
3
)
3
(
5















a
a
a
a
a
a
a
a
Since we obtain the same expression on both sides of the
equation, it is an identity.

aics9e_ppt_2 _1.ppt

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