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Chapter 1
Section 7-1
Linear Equations
7-1-1
© 2008 Pearson Addison-Wesley. All rights reserved
Linear Equations
•
•
•
•
Solving Linear Equations
Special Kinds of Linear Equations
Literal Equations and Formulas
Models
7-1-2
© 2008 Pearson Addison-Wesley. All rights reserved
Equation
An equation is a statement that two
algebraic expressions are equal. A linear
equation in one variable involves only real
numbers and one variable.
4x  7  8 and 5k  3  k 1
Linear equations
7-1-3
© 2008 Pearson Addison-Wesley. All rights reserved
Linear Equation in One Variable
An equation in the variable x is linear if it can
be written in the form
Ax + B = C
where A, B, and C are real numbers, with
A  0.
7-1-4
© 2008 Pearson Addison-Wesley. All rights reserved
Terminology
A linear equation in one variable is also called a firstdegree equation.
If the variable in an equation is replaced by a real
number that makes the statement of the equation true,
then that number is a solution of the equation. An
equation is solved by finding its solution set, the set
of all answers.
Equivalent equations are equations with the same
solution set.
7-1-5
© 2008 Pearson Addison-Wesley. All rights reserved
Solving a Linear Equation in One
Variable
Step 1
Clear fractions. Multiply both
sides by a common denominator.
Step 2
Simplify each side separately.
Clear parentheses and combine
like terms.
Step 3
Isolate the variable terms on one
side. Use the addition property of
equality.
7-1-6
© 2008 Pearson Addison-Wesley. All rights reserved
Solving a Linear Equation in One
Variable
Step 4
Transform so that the coefficient
of the variable is 1. Use the
multiplication property of equality.
Step 5
Check. Substitute the solution
into the original equation.
7-1-7
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Solve a Linear Equation using
the Distributive Property
Solve 3(k + 1) + 2k = 13.
Solution
3k + 3 + 2k = 13
5k + 3 = 13
5k + 3 – 3 = 13 – 3
5k = 10
Distributive property.
Combine like terms.
Subtract 3.
Combine like terms.
7-1-8
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Solve a Linear Equation using
the Distributive Property
Solution (continued)
5k 10

5
5
Divide by 5.
k=2
Check that the solution set is {2} by substituting 2
for k in the original equation.
7-1-9
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Solve a Linear Equation with
Fractions
Solve
x  7 2x  8

 4.
6
2
Solution
 x  7 2x  8 
6

  6  4 
2 
 6
 x  7   2x  8 
6
  6
  6  4 
 6   2 
x  7  3(2 x  8)  24
Multiply by LCD, 6.
Distributive property.
Multiply.
7-1-10
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Solve a Linear Equation with
Fractions
Solution (continued)
x + 7 + 6x – 24 = –24
7x – 17 = –24
7x – 17 + 17 = –24 + 17
7x = –7
7 x 7

7
7
x = –1
Distributive property.
Combine like terms.
Add 17.
Combine like terms.
Divide by 7.
Now check.
7-1-11
© 2008 Pearson Addison-Wesley. All rights reserved
Types of Linear Equations
Type
Conditional
Identity
Number of
Solutions
One
Infinite; solution
set {all real
numbers}
Contradiction None; solution
set 
Final Line When
Solving
x = number
True statement,
such as 0 = 0.
False statement,
such as 0 = 1.
7-1-12
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Contradiction
Solve 3k + 4 – 2k = k.
Solution
k+4=k
k–k+4=k–k
4=0
Combine like terms.
Subtract k.
Combine like terms.
Because the result is false, the equation has no
solution. The solution set is , so the original
equation is a contradiction.
7-1-13
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Identity
Solve 3k + 4 – 2k = k + 4.
Solution
k+4=k+4
k–k+4=k–k
4=4
Combine like terms.
Subtract k.
Combine like terms.
Because the result is true, the solution set is {all real
numbers} and the original equation is an identity.
7-1-14
© 2008 Pearson Addison-Wesley. All rights reserved
Literal Equations and Formulas
An equation involving variables (or letters),
such as y = mx + b is called a literal equation.
Formulas are literal equations in which more
than one letter is used to express a relationship.
It may be necessary to solve for one of the
variables in a formula. This process is called
solving for a specified variable.
7-1-15
© 2008 Pearson Addison-Wesley. All rights reserved
Solving for a Specified Variable
Step 1
Transform the equation so that all
terms containing the specified
variable are on one side of the
equation and all terms without the
variable are on the other side.
Step 2
If necessary, use the distributive
property to combine the terms with
the specified variable.
7-1-16
© 2008 Pearson Addison-Wesley. All rights reserved
Solving for a Specified Variable
Step 3
Divide both sides by the factor that
is multiplied by the specified
variable.
7-1-17
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Solving for a Specified Variable
Solve the formula P = 2L + 2W for L.
Solution
P – 2W = 2L + 2W – 2W
Subtract 2W.
P – 2W = 2L
Combine like terms.
P  2W 2 L

2
2
Divide by 2.
P  2W
P
 L or
 W  L.
2
2
7-1-18
© 2008 Pearson Addison-Wesley. All rights reserved
Models
A mathematical model is an equation (or
inequality) that describes the relationship
between two quantities. A linear model is a
linear equation.
7-1-19
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Temperature Conversion
The relationship between degrees Celsius (C)
and degrees Fahrenheit (F) is modeled by the
linear equation
F = 1.8C + 32.
What Celsius degree corresponds to a
Fahrenheit reading of 50°?
7-1-20
© 2008 Pearson Addison-Wesley. All rights reserved
Example: Temperature Conversion
Solution
Because F = 50, the equation becomes
50 = 1.8C + 32
500 = 18C + 320
Multiply by 10.
180 = 18C
Subtract 320.
C = 10.
Divide by 18.
Therefore, a reading of 50 degrees Fahrenheit
corresponds to a reading of 10 degrees Celsius.
7-1-21
© 2008 Pearson Addison-Wesley. All rights reserved
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