Download Section 1.4 - Warren County 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
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 #3 Section 1.4
Solving Linear Equations
EXAMPLE
Solve and check: 5 - 3x + 4x = 1 - 7x + 12.
SOLUTION
5 - 3x + 4x = 1 - 7x + 12
5 + x = 13 - 7x
5 + x = 13 – 7x
+ 7x
+ 7x
5 + 8x = 13
1) Simplify the algebraic
expressions on each side.
2) Collect variable terms on one
side and constant terms on
the other side.
Blitzer, Algebra for College Students, 6e – Slide #4 Section 1.4
Solving Linear Equations
CONTINUED
5 + 8x = 13
-5
-5
Simplify
Subtract 5 from both sides
8x = 8
8
8
3) Isolate the variable
and solve.
x=1
Divide both sides by 8
Blitzer, Algebra for College Students, 6e – Slide #5 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 #6 Section 1.4
Solving Linear Equations
EXAMPLE #2
Solve and check: 2x – 7 +x = 3x + 1 + 2x.
SOLUTION
Blitzer, Algebra for College Students, 6e – Slide #7 Section 1.4
Solving Linear Equations
EXAMPLE #3
Solve and check: 4(2x + 1) – 29 = 3(2x – 5).
SOLUTION
Blitzer, Algebra for College Students, 6e – Slide #8 Section 1.4
1.4 Assignment (part 1)
p. 46 (2-24 even)

Solving Linear Equations
EXAMPLE #4
Solve and check:
x x
 2
3 2
SOLUTION
x x 2
 
3 2 1

6 x  6 x 2 
    
1 3  1 2 1 
6 x  6 x  6 2 
     
1 3  1 2  1 1 
Multiply both sides
by the LCD: 6
Distributive Property
Blitzer, Algebra for College Students, 6e – Slide #10 Section 1.4


Solving Linear Equations
CONTINUED
6 x  6 x  6 2 
     
1 3  1 2  1 1 
Cancel
2 x  3 x  6 2 
     
1 1  1 1  1 1 
2x  3x  6(2)
2x  3x 12
1x  12
x  12
Multiply
Combine like terms
and solve for x.
Blitzer, Algebra for College Students, 6e – Slide #11 Section 1.4
Solving Linear Equations
EXAMPLE #5
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 #12 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 #13 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 #14 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 #15 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 #16 Section 1.4
Solving Linear Equations
More Practice
EXAMPLE
5 + 4x = 9x + 5
SOLUTION
5 + 4x = 9x + 5
5 - 5 + 4x = 9x + 5 - 5
4x = 9x
4x – 4x = 9x – 4x
0 = 5x
0 5x

5 5
0=x
Subtract 5 from both sides
Simplify
Subtract 4x from both sides
Simplify
Divide both sides by 5
Blitzer, Algebra for College Students, 6e – Slide #17 Section 1.4
1.4 Assignment (part 2)
p. 46 (26-36 even)