Download 2.1 Linear Equations in One Variable

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Sec 2.1 - 1
Chapter 2
Linear Equations and Applications
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Sec 2.1 - 2
2.1
Linear Equations in One Variable
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Sec 2.1 - 3
2.1 Linear Equations in One Variable
Objectives
1.
2.
3.
4.
5.
Decide whether a number is a solution of a linear
equation.
Solve linear equations using the addition and
multiplication properties of equality.
Solve linear equations using the distributive
property.
Solve linear equations with fractions or decimals.
Identify conditional equations, contradictions,
and identities.
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Sec 2.1 - 4
2.1 Using Linear Equations of One Variable
Algebraic Expressions vs. Equations
In the previous chapter, we looked at algebraic
expressions:
– 9y + 5,
10k,
and
a
2 5
- 7b c
Equations are statements that two algebraic
expressions are equal:
3x – 13 = 29, 2 + y = – 11, and 3m = 4m – 2
An equation always contains an equals sign, but an
expression does not.
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Sec 2.1 - 5
2.1 Using Linear Equations of One Variable
Linear Equations in One Variable
Linear Equation in One Variable
A linear equation in one variable can be written in the
form
Ax + B = C
where A, B, and C are real numbers, with A =/ 0.
A linear equation is also called a first-degree
equation since the greatest power on the
variable is one.
5x + 10 = 13
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Sec 2.1 - 6
2.1 Using Linear Equations of One Variable
Linear Equations in One Variable
Determine whether the following equations are
linear or nonlinear.
8x + 3 = –9
Yes, x is raised to the first power.
9x – 8 = 15
No, x is not raised to the first
power.
7 = –12
x
No, x is not raised to the first
power.
3
x  4  16
No, x is not raised to the first
power.
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Sec 2.1 - 7
2.1 Using Linear Equations of One Variable
Deciding Whether a Number is a Solution
If a variable can be replaced by a real number that
makes the equation a true statement, then that
number is a solution of the equation, x – 10 = 3.
13 is a solution
8 is not a solution
13
8
x – 10 = 3
x – 10 = 3
13 – 10 = 3 (true)
8 – 10 = 3
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(false)
Sec 2.1 - 8
2.1 Using Linear Equations of One Variable
Finding the Solution Set of an Equation
An equation is solved by finding its solution set
– the set of all solutions.
The solution set of x – 10 = 3 is {13}.
Equivalent equations are equations that have
the same solution set. These are equivalent
equations since they all have solution set {–3}.
3x + 5 = –4
3x = –9
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x = –3
Sec 2.1 - 9
2.1 Using Linear Equations of One Variable
Solving Linear Equations
An equation is like a balance scale, comparing
the weights of two quantities.
Expression-1
=
Expression-2
We apply properties to produce a series of simpler
equivalent equations to determine the solution set.
Variable
=
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Solution
Sec 2.1 - 10
2.1 Using Linear Equations of One Variable
Addition Property of Equality
The same number may be added to both sides of
an equation without changing the solution set.
A=B
A
=
B
A + C
=
B+C
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Sec 2.1 - 11
2.1 Using Linear Equations of One Variable
Multiplication Property of Equality
Each side of an equation may be multiplied by
the same nonzero number without changing the
solution set.
A=B
A
=
B
AC
=
BC
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Sec 2.1 - 12
2.1 Using Linear Equations of One Variable
Addition and Multiplication Properties of Equality
Addition Property of Equality
For all real numbers A, B, and C, the equation
A = B and A + C = B + C
are equivalent.
Multiplication Property of Equality
For all real numbers A, B, and for C = 0,/ the equation
A = B and A C = B C
are equivalent.
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Sec 2.1 - 13
2.1 Using Linear Equations of One Variable
Addition and Multiplication Properties of Equality
Because subtraction and division are defined in
terms of addition and multiplication,
we can extend the addition and multiplication
properties of equality as follows:
The same number may be subtracted from each side of an
equation, and each side of an equation may be divided by
the same nonzero number, without changing the solution
set.
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Sec 2.1 - 14
2.1 Using Linear Equations of One Variable
Solving Linear Equations in One Variable
Step 1
Clear fractions. Eliminate any fractions by multiplying
each side by the least common denominator.
Step 2
Simplify each side separately. Use the distributive
property to clear parentheses and combine like terms
as needed.
Step 3
Isolate the variable terms on one side. Use the
addition property to get all terms with variables on one
side of the equation and all numbers on the other.
Step 4
Isolate the variable. Use the multiplication property to
get an equation with just the variable (with coefficient of
1) on one side.
Step 5
Check. Substitute the proposed solution into the
original equation.
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Sec 2.1 - 15
2.1 Using Linear Equations of One Variable
Solving Linear Equations
Solve 3x + 2 = 10.
3x + 2 = 10
3x + 2 – 2 = 10 – 2
3x = 8
3x 8

3 3
8
x
3
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Subtract 2.
Combine like terms.
Divide by 3.
Proposed solution.
Sec 2.1 - 16
2.1 Using Linear Equations of One Variable
Solving Linear Equations
3x + 2 = 10
3 • 8 + 2 = 10
3
8 + 2 = 10
Check by substituting the proposed
solution back into the original equation.
Since the value of each side is 10, the
proposed solution is correct.
 8 .
The solution set is  
3
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Sec 2.1 - 17
2.1 Using Linear Equations of One Variable
Solving Linear Equations
Solve 2x – 5 = 5x – 2.
2x – 5 = 5x – 2
2x – 5 – 5x = 5x – 2 – 5x
–3x – 5 = –2
–3x – 5 + 5= –2 + 5
–3x = 3
3 x 3

3
3
x = –1
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Subtract 5x.
Combine like terms.
Add 5.
Combine like terms.
Divide by –3.
Proposed solution.
Sec 2.1 - 18
2.1 Using Linear Equations of One Variable
Solving Linear Equations
2x – 5 = 5x – 2
Check by substituting the
proposed solution back into
the original equation.
2(–1) – 5 = 5(–1) – 2
–2 – 5 = –5 – 2
Since the value of each
side is –7 , the proposed
solution is correct.
–7 = –7
The solution set is {–1}.
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Sec 2.1 - 19
2.1 Using Linear Equations of One Variable
Solving Linear Equations
Solve 5(2x + 3) = 3 – 2(3x – 5).
5(2x + 3) = 3 – 2(3x – 5)
10x + 15 = 3 – 6x + 10
Distributive Prop.
10x + 15 – 15 = 3 – 6x + 10 – 15
Add –15.
10x = – 6x – 2
Collect like terms.
10x + 6x = –6x – 2 + 6x
Add 6x.
16x = –2
Collect like terms.
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Sec 2.1 - 20
2.1 Using Linear Equations of One Variable
Solving Linear Equations
16x = –2
16 x 2

16 16
1
x
8
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Divide by 16.
Proposed solution.
Sec 2.1 - 21
2.1 Using Linear Equations of One Variable
Solving Linear Equations
Check proposed solution:
5  2 x  3  3  2  3 x  5 
  1 
  1 
5  2     3   3  2 3     5 
  8 
  8 
 2 
 3 
5    3  3  2    5 
 8 
 8 
 2 24 
 3 40 
5     3  2   
 8 8
 8 8
 1
The solution set set is   .
 8
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 22 
 43 
5   3  2  
 8 
 8 
110
86
 3
8
8
110 24 86


8
8
8
110 110

Checks
8
8
Sec 2.1 - 22
2.1 Using Linear Equations of One Variable
Solving Linear Equations with Fractions
Solve
2x 1 1 x  3
 
2
3
4
 2x 1 
1 x 3
12 
  12  

2
3
4




6  2 x  1  4  3  x  3
12 x  6  4  3 x  9
12 x  6  3 x  4  3 x  9  3x
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.
Clear fractions.
Distributive property.
Distributive property.
Add  3 x.
Sec 2.1 - 23
2.1 Using Linear Equations of One Variable
Solving Linear Equations with Fractions
continued
12 x  6  x  4  3 x  9  3 x
9 x  6  5
9 x  6  6  5  6
9 x  1 1
9 x 11

9
9
11
x
9
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Collect like terms.
Add  6.
Collect like terms
Divide by 9.
Proposed solution.
Sec 2.1 - 24
2.1 Using Linear Equations of One Variable
Solving Linear Equations with Decimals
Solve 1.5 ( x + 2) = 2.8 + x
1.5  x  2   2.8  x
15  x  2   28  10 x
15 x  30  28  10 x
15 x  30  10 x  28  10 x  10 x
5 x  30  28
5 x  30  30  28  30
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.
Multiply by 10.
Distributive property.
Add  10 x.
Collect like terms.
Add  30.
Sec 2.1 - 25
2.1 Using Linear Equations of One Variable
Solving Linear Equations with Decimals
continued
5 x  30  30  28  30
5 x  2
5 x 2

5
5
2
x
5
Collect like terms.
Divide by 5.
Proposed solution.
 2
The solution set is   .
 5
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Sec 2.1 - 26
2.1 Using Linear Equations of One Variable
Conditional, Contradiction, and Identity Equations
Linear equations can have exactly one solution,
no solution, or an infinite number of solutions.
Type of Linear
Equation
Number of Solutions
Indication When
Solving
Conditional
One
Final results is
x = a number.
Identity
Infinite; solution set
{all real numbers}
Final line is true,
such as 5 = 5.
Contradiction
None; solution set is
Final line is false,
such as –3 = 11.
.
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Sec 2.1 - 27
2.1 Using Linear Equations of One Variable
Conditional, Contradiction, and Identity Equations
A contradiction has no solutions.
Solve x  7  x  2.
x7  x2
x77  x27
x  x5
x  x  x 5 x
0  5
Adding  7.
Collecting like terms.
Add  x.
Collecting like terms.
Since 0 = –5 is never true, and this equation is equivalent
to x + 7 = x + 2, the solution set is empty.
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Sec 2.1 - 28
2.1 Using Linear Equations of One Variable
Conditional, Contradiction, and Identity Equations
An identity has an infinite number of solutions.
Solve 2x  2  2  x  1 .
2 x  2  2  x  1
2x  2  2x  2
Distributive property.
2x  2  2  2x  2  2
Adding  2.
2x  2x
Collecting like terms.
2x  2x  2x  2x
00
Adding  2 x.
Collecting like terms.
Since 0 = 0 is always true, and this equation is equivalent
to 2x + 2 = 2(x + 1), the solution set is all real numbers.
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Sec 2.1 - 29