Linear equations

1. Introduction to Mathematics
Course Code: ENG (GC) 104
Linear Equations

2. •Equations and Inequalities
1

3. Overview
• Equations are the basic mathematical tool for solving real-world
problems.
• In this chapter , we learn
• How to solve equations
• How to construct equations that real-life situations.

4. •Basic Equations
1.1

5. Equations
• An equation is a statement that two
mathematical expressions are equal.
• For example:
3 + 5 = 8

6. Variables
• Most equations that we study in algebra
contain variables, which are symbols (usually
letters) that stand for numbers.
• In the equation 4x + 7 = 19, the letter x
is the variable.
• We think of x as the “unknown” in the equation.
• Our goal is to find the value of x that makes
the equation true.

7. Solving the Equation
• The values of the unknown that make
the equation true are called the solutions or
roots of the equation.
• The process of finding the solutions is called solving
the equation.

8. Equivalent Equations
• Two equations with exactly the same solutions
are called equivalent equations.
• To solve an equation, we try to find a simpler, equivalent equation in which
the variable stands alone on one side of the “equal” sign.

9. Properties of Equality
• We use the following properties to solve
an equation.
• A, B, and C stand for any algebraic expressions.
• The symbol means “is equivalent to.”


10. Properties of Equality
• These properties require that you perform
the same operation on both sides of an
equation when solving it.
• So, if we say “add –7” when solving an equation, that is just a short way of
saying “add –7 to each side of the equation.”

11. •Solving Linear Equations

12. Linear Equation
• The simplest type of equation is a linear
equation, or first-degree equation.
• It is an equation in which each term is either
a constant or a nonzero multiple of the variable.

13. Linear Equation—Definition
• A linear equation in one variable is an equation equivalent to one
of the form
ax + b = 0
• where:
• a and b are real numbers.
• x is the variable.

14. Linear Equations
• These examples illustrate the difference
between linear and nonlinear equations.
Linear
Equations
Nonlinear
Equations
Reason for
Being Nonlinear
4x – 5 = 3 x2 + 2x = 8 Contains the square
of the variable
2x = ½x – 7 Contains the square root
of the variable
x – 6 = x/3 3/x – 2x = 1 Contains the reciprocal
of the variable
 
6 0
x x

15. E.g. 1 - Solving a Linear Equation
•Solve the equation
7x – 4 = 3x + 8
• We solve this by changing it to an equivalent equation with all terms that
have the variable x on one side and all constant terms on the other.

16. E.g. 1 - Solving a Linear Equation
• 7x – 4 = 3x + 8
• (7x – 4) + 4 = (3x + 8) + 4 (Add 4)
• 7x = 3x + 12 (Simplify)
• 7x – 3x = (3x + 12) – 3x (Subtract 3x)
• 4x = 12 (Simplify)
• ¼ . 4x = ¼ . 12 (Multiply by ¼)
• x = 3 (Simplify)

17. Checking Your Answer
• It is important to check your answer.
• We do so in many examples.
• In these checks, LHS stands for “left-hand side” and RHS stands for “right-
hand side” of the original equation.

18. Checking Your Answer
• We check the answer of Example 1.
• x = 3:
• LHS = 7(3) – 4 = 17
• RHS = 3(3) + 8 = 17
• LHS = RHS

19. E.g. 2 - Solving an Equation Involving
Fractions
•Solve the equation
• The LCD of the denominators 6, 3, and 4 is 12.
• So, we first multiply each side of the equation by 12 to clear denominators.
 
2 3
6 3 4
x
x

20. E.g. 2 - Solving an Equation Involving
Fractions
 
 
 
 
 


2 3
12 12
6 3 4
2 8 9
8 7
8
7
x
x
x x
x
x

21. An Equation that Simplifies to a Linear Equation
• In the next example, we solve an equation the doesn’t look like a
linear equation, but it simplifies to one when we multiply by the
LCD.

22. E.g. 3 - Equation Involving Fractional
Expressions
•Solve the equation
• The LCD of the fractional expressions is
(x + 1)(x – 2) = x2 – x – 2.
• So, as long as x ≠ –1 and x ≠ 2, we can multiply both sides by the LCD.

 
   
2
1 1 3
1 2 2
x
x x x x

23. E.g. 3 - Equation Involving Fractional
Expressions

   
     
   
   
   
    
  

2
1 1 3
( 1)( 2) ( 1)( 2)
1 2 2
( 2) ( 1) 3
2 1 3
4
x
x x x x
x x x x
x x x
x x
x

24. Extraneous Solutions
• It is always important to check your answer.
• Even if you never make a mistake in your calculations.
• This is because you sometimes end up with extraneous solutions.

25. Extraneous Solutions
• Extraneous solutions are potential solutions that do not satisfy the
original equation.
• The next example shows how this can happen.

26. E.g. 4 - An Equation with No Solution
•Solve the equation
• First, we multiply each side by the common denominator, which is x – 4.

 
 
5 1
2
4 4
x
x x

27. E.g. 4 - An Equation with No Solution

   
   
   
 
   
   
   
  
 

5 1
( 4) 2 ( 4)
4 4
2( 4) 5 1
2 8 5 1
2 3 1
2 4
4
x
x x
x x
x x
x x
x x
x x
x

28. E.g. 4 - An Equation with No Solution
•Now, we try to substitute x = 4 back into
the original equation.
• We would be dividing by 0, which is impossible.
• So this equation has no solution.

29. Extraneous Solutions
• The first step in the preceding solution, multiplying by x – 4, had the
effect of multiplying by 0.
• Do you see why?
• Multiplying each side of an equation by an expression that contains the variable
may introduce extraneous solutions.
• That is why it is important to check every answer.

30. • Solving Power Equations

31. Power Equations
• Linear equations have variables only to the
first power.
• Now let’s consider some equations that involve
squares, cubes, and other powers of the
variable.
• Such equations will be studied more extensively in Sections 1.3 and 1.5.

32. Power Equation
• Here we just consider basic equations that can
be simplified into the form
Xn = a
• Equations of this form are called power equations
• They are solved by taking radicals of both sides to
the equation.

33. Solving a Power Equation
• The power equation Xn = a has the solution
• X = if n is odd
• X = if n is even and a ≥ 0
• If n is even and a < 0, the equation has no real
solution.
n
a
n
a

34. Examples of Solving a Power
Equation
• Here are some examples of solving power
equations.
• The equation x5 = 32 has only one real solution: x =
= 2.
• The equation x4 = 16 has two real solutions: x =
= ±2.
5
32
4
16

35. Examples of Solving a Power
Equation
• Here are some examples of solving power
equations.
• The equation x5 = –32 has only one real solution:
x = = –2.
• The equation x4 = –16 has no real solutions
because does not exist.

5
32

4
16

36. E.g. 5 - Solving Power Equations
• Solve each equation.
(a) x2 – 5 = 0
(b) (x – 4)2 = 5

37. E.g. 5—Solving Power Equations
•x2 – 5 = 0
• x2 = –5
• x = ±
•The solutions are x = and x = .
Example (a)
5
5
5

38. E.g. 5—Solving Power Equations
• We can take the square root of each side
of this equation as well.
• (x – 4)2 = 5
• x – 4 = ±
• x = 4 ± (Add 4)
5
5
Example (b)

39. E.g. 6—Solving Power Equations
• Find all real solutions for each equation.
(a) x3 = – 8
(b) 16x4 = 81

40. E.g. 6—Solving Power Equations
•Since each real number has exactly one
real cube root, we can solve this
equation by taking the cube root of
each side.
• (x3)1/3 = (–8)1/3
• x = –2
Example (a)

41. E.g. 6—Solving Power Equations
• Here we must remember that if n is even, then every positive
real number has two real nth roots.
• A positive one and a negative one.
• If n is even, the equation xn = c (c > 0) has two solutions, x = c1/n and
x = –c1/n.
Example (b)

42. E.g. 6—Solving Power Equations

 
  
 
 
4
1/ 4
4 1/ 4
81
16
81
( )
16
3
2
x
x
x
Example (b)

43. Equations with Fractional Power
• The next example shows how to solve an equation that involves a
fractional power of the variable.

44. E.g. 7—An Equation with a Fractional Power
• Solve the equation
• 5x2/3 – 2 = 43
• The idea is to first isolate the term with the fractional exponent, then raise
both sides of the equation to the reciprocal of that exponent.
• If n is even, the equation xn/m = c has two solutions, x = cm/n and x = –cm/n.

45. E.g. 7—An Equation with a
Fractional Power
• 5x2/3 – 2 = 43
• 5x2/3 = 45
• x2/3 = 9
• x = ±93/2
• x = ±27

46. Complex Numbers and Roots
You can see in the graph of f(x) = x2 + 1 below
that f has no real zeros. If you solve the
corresponding equation 0 = x2 + 1, you find
that x = ,which has no real solutions.
However, you can find solutions if you
define the square root of negative
numbers, which is why imaginary
numbers were invented. The
imaginary unit i is defined
as . You can use the imaginary
unit to write the square root of
any negative number.

47. Complex Numbers and Roots

48. Express the number in terms of i.
Example 1A: Simplifying Square Roots of Negative
Numbers
Factor out –1.
Product Property.
Simplify.
Multiply.
Express in terms of i.

49. Express the number in terms of i.
Example 1B: Simplifying Square Roots of Negative
Numbers
Factor out –1.
Product Property.
Simplify.
Express in terms of i.
4 6 4 6
i i
  

50. Huddle
Express the number in terms of i.
Factor out –1.
Product Property.
Simplify.
Express in terms of i.
Product Property.

51. Huddle
Express the number in terms of i.
Factor out –1.
Product Property.
Simplify.
Express in terms of i.
Multiply.

52. Mastery
Express the number in terms of i.
Factor out –1.
Product Property.
Simplify.
Express in terms of i.
Multiply.

53. Solve the equation.
Example 2A: Solving a Quadratic Equation with
Imaginary Solutions
Take square roots.
Express in terms of i.
Check
x2 = –144
–144
–144
–144
(12i)2
144i 2
144(–1) 
x2 = –144
–144
–144
–144
144(–1)
144i 2
(–12i)2


54. Solve the equation.
Example 2B: Solving a Quadratic Equation with
Imaginary Solutions
Add –90 to both sides.
Divide both sides by 5.
Take square roots.
Express in terms of i.
Check 5x2 + 90 = 0
0
0
0
5(18)i 2 +90
90(–1) +90 
5x2 + 90 = 0

55. Huddle
x2 = –36
Solve the equation.
Take square roots.
Express in terms of i.
x2 = –36
–36
–36
–36
(6i)2
36i 2
36(–1) 
x2 = –36
–36
–36
–36
(–6i)2
36i 2
36(–1) 
Check

56. Huddle
x2 + 48 = 0
Solve the equation.
Add –48 to both sides.
Take square roots.
Express in terms of i.
x2 = –48
Check x2 + 48 = 0
0
0
0
(48)i 2 + 48

+ 48
48(–1) + 48

57. Mastery
9x2 + 25 = 0
Solve the equation.
Add –25 to both sides.
Divide both sides by 9.
Take square roots.
Express in terms of i.
9x2 = –25

58. Every complex number has a real part a and an
imaginary part b.
A complex number is a
number that can be written
in the form a + bi, where a
and b are real numbers and
i = . The set of real
numbers is a subset of the
set of complex numbers C.
Complex Numbers and Roots

59. Find the zeros of the function.
Example 4A: Finding Complex Zeros of Quadratic
Functions
x2 + 10x + = –26 +
f(x) = x2 + 10x + 26
Add to both sides.
x2 + 10x + 26 = 0 Set equal to 0.
Rewrite.
x2 + 10x + 25 = –26 + 25
(x + 5)2 = –1 Factor.
Take square roots.
Simplify.

60. Find the zeros of the function.
Example 4B: Finding Complex Zeros of Quadratic
Functions
Add to both sides.
g(x) = x2 + 4x + 12
x2 + 4x + 12 = 0
x2 + 4x + = –12 +
x2 + 4x + 4 = –12 + 4
Rewrite.
Set equal to 0.
Take square roots.
Simplify.
Factor.
(x + 2)2 = –8

61. Huddle
Find the zeros of the function.
x2 + 4x + = –13 +
f(x) = x2 + 4x + 13
Add to both sides.
x2 + 4x + 13 = 0 Set equal to 0.
Rewrite.
x2 + 4x + 4 = –13 + 4
(x + 2)2 = –9 Factor.
Take square roots.
Simplify.
x = –2 ± 3i

62. Mastery
Find the zeros of the function.
x2 – 8x + = –18 +
g(x) = x2 – 8x + 18
Add to both sides.
x2 – 8x + 18 = 0 Set equal to 0.
Rewrite.
x2 – 8x + 16 = –18 + 16
Factor.
Take square roots.
Simplify.

63. If a quadratic equation with real coefficients has
nonreal roots, those roots are complex conjugates.
When given one complex root, you can always
find the other by finding its conjugate.
Helpful Hint
The solutions and are related.
These solutions are a complex conjugate pair.
Their real parts are equal and their imaginary
parts are opposites. The complex conjugate of
any complex number a + bi is the complex
number a – bi.

64. Find each complex conjugate.
Example 5: Finding Complex Zeros of Quadratic
Functions
A. 8 + 5i B. 6i
0 + 6i
8 + 5i
8 – 5i
Write as a + bi.
Find a – bi.
Write as a + bi.
Find a – bi.
Simplify.
0 – 6i
–6i

65. Find each complex conjugate.
A. 9 – i
9 + (–i)
9 – (–i)
Write as a + bi.
Find a – bi.
Huddle
B.
C. –8i
0 + (–8)i Write as a + bi.
Find a – bi.
Simplify.
0 – (–8)i
8i
Write as a + bi.
Find a – bi.
9 + i Simplify.

66. Home Assignment
Solve each equation.
2. 3x2 + 96 = 0 3. x2 + 8x +20 = 0
1. Express in terms of i.
4. Find the values of x and y that make the
equation 3x +8i = 12 – (12y)i true.
5. Find the complex conjugate of