Download 1.4 Solving Linear Equations - Warren County Public Schools

§ 1.4
Solving Linear Equations
Linear Equations
Definition of a Linear Equation
A linear equation in one variable x is an
equation that can be written in the form
ax + b = 0, where a and b are real numbers and
a is not equal to 0.
An example of a linear equation in x is 4x + 2 = 6. Linear
equations in x are first degree equations in the variable x.
Blitzer, Algebra for College Students, 6e – Slide #2 Section 1.4
Properties of Equality
Property
Definition
Addition Property of Equality
The same real number or algebraic
expression may be added to both sides
of an equation without changing the
equation’s solution set.
Multiplication Property of Equality
The same nonzero real number may
multiply both sides of an equation
without changing the equation’s
solution set.
Blitzer, Algebra for College Students, 6e – Slide #3 Section 1.4
Solving Linear Equations
Solving a Linear Equation
1) Simplify the algebraic expressions on each side.
2) Collect all the variable terms on one side and all the
numbers, or constant terms, on the other side
3) Isolate the variable and solve.
4) Check the proposed solution in the original equation.
Blitzer, Algebra for College Students, 6e – Slide #4 Section 1.4
Solving Linear Equations
EXAMPLE
Solve and check: 5 - 3x + 4x = 1 - 7x + 12.
SOLUTION
1) Simplify the algebraic expressions on each side.
5 - 3x + 4x = 1 - 7x + 12
5 + x = 13 - 7x
Combine like terms:
-3x + 4x = x
1 + 12 = 13
Blitzer, Algebra for College Students, 6e – Slide #5 Section 1.4
Solving Linear Equations
CONTINUED
2) Collect variable terms on one side and constant terms
on the other side.
5 + x + 7x = 13 - 7x + 7x
5 + 8x = 13
5 – 5 + 8x = 13 - 5
8x = 8
Add 7x to both sides
Simplify
Subtract 5 from both sides
Simplify
Blitzer, Algebra for College Students, 6e – Slide #6 Section 1.4
Solving Linear Equations
CONTINUED
3) Isolate the variable and solve.
8x  8
8
8
Divide both sides by 8
x=1
Simplify
Blitzer, Algebra for College Students, 6e – Slide #7 Section 1.4
Solving Linear Equations
CONTINUED
4) Check the proposed solution in the original equation.
5 - 3x + 4x = 2 - 7x + 6
5 – 3(1) + 4(1) ?= 1 – 7(1) + 12
5 – 3 + 4 ?= 1 – 7 + 12
2 + 4 ?= – 6 + 12
6=6
Original equation
Replace x with 1
Multiply
Add or subtract from left to
right
Add
Blitzer, Algebra for College Students, 6e – Slide #8 Section 1.4
Solving Linear Equations
EXAMPLE
Solve and check:
SOLUTION
2x 1 x  2 x


3
5
2
2x 1 x  2 x


3
5
2
1) Simplify the algebraic expressions on each side.
 2x  1 x  2 
 x
30

  30 
5 
 3
2
30  2 x  1  30  x  2  30  x 

 
  
1  3  1  5  1 2
Multiply both sides
by the LCD: 30
Distributive Property
Blitzer, Algebra for College Students, 6e – Slide #9 Section 1.4
Solving Linear Equations
CONTINUED
30  2 x  1  30  x  2  30  x 

 
  
1  3  1  5  1 2
10  2 x  1  6  x  2  15  x 

 
  
1  1  1 1  1 1
102x 1  6x  2  15x
20x 10  6x 12  15x
Cancel
Multiply
Distribute
Blitzer, Algebra for College Students, 6e – Slide #10 Section 1.4
Solving Linear Equations
CONTINUED
14x + 2 = 15x
Combine like terms
2) Collect variable terms on one side and constant terms
on the other side.
14x – 14x + 2 = 15x – 14x
2=x
Subtract 14x from both sides
Simplify
3) Isolate the variable and solve.
Already done.
Blitzer, Algebra for College Students, 6e – Slide #11 Section 1.4
Solving Linear Equations
CONTINUED
4) Check the proposed solution in the original equation.
2x 1 x  2 ? x


3
5
2
Original Equation
22   1 2  2 ? 2


3
5
2
Replace x with 2
4 1 2  2 ? 2


3
5
2
Simplify
3 0 ?2
 
3 5 2
Simplify
Blitzer, Algebra for College Students, 6e – Slide #12 Section 1.4
Solving Linear Equations
CONTINUED
1-0=1
Simplify
1=1
Simplify
Since the proposed x value of 2 made a true sentence
of 1 = 1 when substituted into the original equation,
then 2 is indeed a solution of the original equation.
Blitzer, Algebra for College Students, 6e – Slide #13 Section 1.4
Categorizing an Equations
Type of Equations
Identity
Conditional
Inconsistent
(contradiction)
Definitions
An equation that is true for
all real numbers
An equation that is not an
identity but is true for at
least one real number
An equation that is not true
for any real number
Blitzer, Algebra for College Students, 6e – Slide #14 Section 1.4
Categorizing an Equation
EXAMPLE
Solve and determine whether the equation is an identity,
a conditional equation or an inconsistent equation.
5 + 4x = 9x + 5
SOLUTION
5 + 4x = 9x + 5
5 - 5 + 4x = 9x + 5 - 5
4x = 9x
4x – 4x = 9x – 4x
Subtract 5 from both sides
Simplify
Subtract 4x from both sides
Blitzer, Algebra for College Students, 6e – Slide #15 Section 1.4
Categorizing an Equation
CONTINUED
0 = 5x
Simplify
0 5x

5 5
Divide both sides by 5
0=x
Simplify
The original equation is only true when x = 0.
Therefore, it is a conditional equation.
Blitzer, Algebra for College Students, 6e – Slide #16 Section 1.4
Categorizing an Equation
EXAMPLE
Solve and determine whether the equation is an identity,
a conditional equation or an inconsistent equation.
5 – (2x – 4) = 4(x +1) - 2x
SOLUTION
5 – (2x – 4) = 4(x +1) - 2x
5 – 2x +4 = 4x + 4 -2x
9 - 2x = 4 - 2x
9=4
Distribute the -1 and the 4
Simplify
Add 2x to both sides.
Since after simplification we see a contradiction, we know that the
original equation is inconsistent and can never be true for any x.
Blitzer, Algebra for College Students, 6e – Slide #17 Section 1.4
Categorizing an Equation
EXAMPLE
Solve and determine whether the equation is an identity,
a conditional equation or an inconsistent equation.
3 + 2x = 3(x +1) - x
SOLUTION
3 + 2x = 3(x +1) - x
3 + 2x = 3x + 3 - x
Distribute the 3
3 + 2x = 2x + 3
Simplify
Since after simplification we can see that the left hand side (LHS) is equal to
the RHS of the equation, this is an identity and is always true for all x.
Blitzer, Algebra for College Students, 6e – Slide #18 Section 1.4