Transcript PPT Chapter 03 - McGraw Hill Higher Education

Introductory Mathematics
& Statistics
Chapter 3
Algebra
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Learning Objectives
• Understand and use algebraic terms
• Manipulate algebraic expressions
• Solve simple linear equations (using transposition)
• Solve simultaneous linear equations
• Solve business problems using simple algebra
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3.1 Definitions
• Algebra applies quantitative concepts to unknown quantities
represented by symbols
• A constant is a term whose value does not change
E.g. 12, 3.5, 5 , 
8
• A variable is a term that represents a quantity that may have
different values
E.g. x, y, z
• An (algebraic) expression is a combination of constants and
variables using arithmetic operations
E.g.
x 3  3z2  5xy
xy
2x  3y
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3.1 Definitions (cont…)
• A term is part of an expression that is connected to another
term by a + or a – sign
E.g. In the expression 2x + 6y – 4z, the terms are 2x, 6y
and –4z
• A coefficient is a factor by which the rest of a term is
multiplied
E.g. The term 9xy has a coefficient of 9
• The degree of expression is the highest exponent of any
variable in the expression
E.g. the expression 3x2 + 9x + 5 is a quadratic or second
degree expression
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3.1 Definitions (cont…)
• An equation is a statement that two expressions are
equal
• A linear equation is one in which the largest value of
the exponents is 1
• Solving the equation is where we find the value for the
variable which makes the equation a true statement
• Two simultaneous equations must be solved at the
same time to find the values of the variables that will
satisfy both equations
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3.1 Definitions (cont…)
• A formula is a rule or principle that is expressed in terms
of algebraic symbols
E.g. Formula to find the area of a circle
A  r 2
• Formulas can be rewritten to make another variable the
subject
E.g.
r
A

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3.2 Algebraic expressions
• Rule:
If an expression contains like terms, these terms may be
combined into a single term. Like terms are terms that differ
only in their numerical coefficient. Constants may also be
combined into a single constant
Example:
Simplify the exp ression 5x  3x  4z
5 x and 3 x are like terms
5 x and - 3 x may be combined
to form a single term
5 x - 3x  2 x
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3.2 Algebraic expressions (cont…)
• Rule:
When an expression is contained in brackets, each term
within the brackets is multiplied by any coefficient outside
the brackets. This is called expanding
Example:
Consider the expression :
23 x  4 y  1
to remove the brackets
23 x   24 y   21  6 x  8 y  2
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3.2 Algebraic expressions (cont…)
• Rule:
To multiply one expression by another, multiply each term of
one expression by each term of the other expression. The
resulting expression is said to be the product of the two
expressions.
Example:
The product of the two expressions
3x  2 and 2x  1
3x  22x  1  3x 2x  1  22x  1
 6x 2  3x  4x  2
 6x 2  x  2
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3.3 Solving linear equations
•
When solving an equation that involves only one
variable, follow these steps:
1. Place all the terms involving the variable on the left-hand
side of the equation and the constant terms on the righthand side
2. Collect the like terms, and collect the constant terms
3. Divide both sides of the equation by the coefficient of the
variable
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3.3 Solving linear equations (cont…)
• The left-hand side of the equation should now
consist of the variable only. The right-hand side of
the equation is the solution
• When terms are moved from one side of an
equation to the other, they are said to be
transposed (or transferred). This process is called
transposition
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3.3 Solving linear equations (cont…)
• For example:
Solve
9x – 27 = 4x + 3 for x
1. Place like terms of the variable on the left side of the equation
and the constant terms on the right side
9x – 4x = 3 + 27
2. Collect like terms and constant terms
5x = 30
3. Divide both sides of the equation by the coefficient of the
variable (in this case 5).
x=6
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3.3 Solving linear equations (cont…)
• Rule:
Any term may be transposed from one side of an equation
to the other. When the transposition is made, the sign of
the term must change from its original sign. That is, a ‘+’
becomes a ‘-’ and a ‘-’ becomes a ‘+’.
Example:
15x - 20 = 12 - 4x
15x - 20 + 4x = 12
15x + 4x = 12 + 20
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3.4 Solving simultaneous linear
equations
• We consider the situation in which we have to find the values
of two variables
• We must have two equations involving the variables
• Such equations are referred to as simultaneous equations
• In this case, there will usually be unique values of the
variables that will satisfy both equations
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3.4 Solving simultaneous linear
equations (cont...)
• Solve for x and y
3x + 4y = 33
2x – 3y = 5
(1)
(2)
• Step 1:
Make the coefficient of either of the variables in one equation
equal to its coefficient in the other equation. Multiply both
sides of equation (1) by 2 and equation (2) by 3.
6x + 8y = 66
6x – 9y = 15
(3)
(4)
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3.4 Solving simultaneous linear
equations (cont...)
• Step 2:
Eliminate the variable that has the same coefficient by:
subtracting equation (4) from equation (3)
–
If the signs are the same, then subtract equations
–
If the signs are different, then add equations
6x + 8y = 66 (3)
minus: 6x – 9y = 15 (4)
equals 8y – (– 9y) = 51
17y = 51
Divide both sides by 17 to find y
Therefore:
y = 3
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3.4 Solving simultaneous linear
equations (cont...)
•
Substitute in 3 for y in equation (1) and solve for x.
3x + 4(3) = 33
3x + 12 = 33
3x
= 33 - 12
3x
= 21
Divide both sides by 3
Therefore:
x=7
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3.5 Applications
• There are many problems in which the value of some
unknown quantity is to be found
• In many cases, we can represent the unknown quantity by
some variable name and construct an equation involving that
variable
• There is a vast array of such problems, and the best way to
describe them is to look at different examples and then
determine how they are solved
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3.5 Applications (cont…)
Example
Tran left his car for service and received a combined bill (for parts and
labour) of $228. Given that the labour costs twice as much as the
parts, find the amount that parts and labour each cost
Solution
Let the amount charged for parts (in dollars) = x
Then the amount charged for labour (in dollars) = 2x
x  2x
3x
3x
3
x
 $228
 228
228

3
 76
Therefore, the cost of parts = $x = $76
and the cost of labour = $2x = $2 ( 76 ) = $152
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PowerPoint slides to accompany Croucher, Introductory Mathematics and Statistics, 5e
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Summary
•
We have looked at different algebraic definitions
•
We have also looked at simplifying algebraic expressions
•
We examined solving simple linear equations, which is solving
for one variable
•
We also solved simultaneous linear equations, which is solving
equations with two variables
•
Lastly we had a brief look at the applications of algebra
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