When we play games with computers we play by running, jumping and or finding secret things. Well, with Algebra we play with letters, numbers and symbols. And we also get to find secret things. Once we learn some of the ‘tricks’ it becomes a fun challenge to work with our skills in solving each puzzle. So, Algebra is all about solving puzzles. In this
chapter we are going to learn some of the skills that help in solving mathematics puzzles.
chapter we are going to learn some of the skills that help in solving mathematics puzzles.
The Binary Operations
Describe the binary operations
When two numbers are combined according to the instructions given and produce one number we say that they are binary operations. For example, when we add 4 to 6 we get 10, or when we multiply 4 by 3 we get 12. We see that addition of two numbers lead to one number and multiplication of two numbers produce one number. This is binary operation. The instructions can be given either by symbols like X,*,∇and so on or by
words.
words.
Performing Binary Operations
Perform binary operations
Example 1
Evaluate:
solution
Example 2
Find
solution
Example 3
Solve
Example 4
evaluate,
Example 5
Calculate
Brackets in Computation
Brackets are used to group items into brackets and these items inside the brackets are considered as whole. For example,15 ÷(X + 2) ,means
that x and 2 are added first and their sum should divide 15. If we are
given expression with mixed operations, the following order is used to
perform the operations: Brackets (B) are opened (O) first followed by
Division (D) then Multiplication (M), Addition (A) and lastly
Subtraction (S). Shortly is written as BODMAS.
that x and 2 are added first and their sum should divide 15. If we are
given expression with mixed operations, the following order is used to
perform the operations: Brackets (B) are opened (O) first followed by
Division (D) then Multiplication (M), Addition (A) and lastly
Subtraction (S). Shortly is written as BODMAS.
Basic Operations Involving Brackets
Perform basic operations involving brackets
Example 6
Simplify the following expressions:
- 4 + 2b – (9b ÷3b)
- 4z – (2x + z)
solution
Algebraic Expressions Involving the Basic Operations and Brackets
Simplify algebraic expressions involving the basic operations and brackets
Example 7
Evaluate the following expressions:
solution
Identities
For
example, 3(2y + 3) = 6y + 9, when y = 1, the right hand side (RHS) and
the left hand side (LHS) are both equals to 15. If we substitute any
other, we obtain the same value on both sides. Therefore the equations
which are true for all values of the variables on both sides are called
Identities. We can determine whether an equation is an identity or not
by showing that an expression on one side is identical to the other
expression on the other side.
example, 3(2y + 3) = 6y + 9, when y = 1, the right hand side (RHS) and
the left hand side (LHS) are both equals to 15. If we substitute any
other, we obtain the same value on both sides. Therefore the equations
which are true for all values of the variables on both sides are called
Identities. We can determine whether an equation is an identity or not
by showing that an expression on one side is identical to the other
expression on the other side.
Example 8
Determine whether or not the following expressions are identities:
solution
Quadratic Expressions
A Quadratic Expression from Two Linear Factors
Form a quadratic expression from two linear factors
A quadratic expression is an expression where the highest exponent of the variable (usually x) is a square (x2). It is usually written as ax2+bx+c.
Activity 1
Form a quadratic expression from two linear factors
The General Form of Quadratic Expression
Write the general form of quadratic expression
Quadratic expression has the general form of ax2 + bx + c where a ≠ 0 and a is a coefficient of x2 , b is a coefficient of x and c is a constant. its highest power of variable is 2. Examples of quadratic expressions are 2x2 + x + 1, 4y2 + 3, 3z2 – 4z + 1 and so on. In a quadratic expression 3z2 – 4z + 1, a = 3, b = -4 and c = 1. Also in quadratic expression 4y2 + 3, a = 4, b = 0 and c = 3
Example 9
If you are told to find the area of a rectangle with a length of 4y + 3 and a width of 2y + 1.
Solution
Example 10
3x items were bought and each item costs (4x – 3) shillings. Find total amount of money used.
Linear Expressions
Factorize linear expressions
The
operation of resolving a quantity into factors, when we expand
expressions, is done by removing the brackets. The reverse operation is
Factorizing and it is done by adding brackets.
operation of resolving a quantity into factors, when we expand
expressions, is done by removing the brackets. The reverse operation is
Factorizing and it is done by adding brackets.
Example 11
Factorize the expression 5a+5b.
Solution
In
factorization of 5a+5b, we have to find out a common thing in both
terms. We can see that the expression 5a+5b, have got common coefficient
in both terms, that is 5. So factoring it out we get 5(a+b).
factorization of 5a+5b, we have to find out a common thing in both
terms. We can see that the expression 5a+5b, have got common coefficient
in both terms, that is 5. So factoring it out we get 5(a+b).
Example 12
Factorize 18xyz-24xwz
Solution
Factorizing
18xyz-24xwz, we have to find out highest common factor of both terms.
Then factor it out, the answer will be 9xz(2y-3w).
18xyz-24xwz, we have to find out highest common factor of both terms.
Then factor it out, the answer will be 9xz(2y-3w).
Quadratic Expressions
Factorize quadratic expressions
When
we write the quadratic expression as a product of two factors we say
that we have factorized the expression. We are going to learn two
methods used to factorize quadratic expressions. These methods are
factorization by Splitting the middle term and factorization by
Inspection.
we write the quadratic expression as a product of two factors we say
that we have factorized the expression. We are going to learn two
methods used to factorize quadratic expressions. These methods are
factorization by Splitting the middle term and factorization by
Inspection.
Factorization by splitting the middle term
Example 13
factorize 3x2 – 2x – 8 by splitting the middle term.
Solution
Example 14
factorize x2 + 10x + 25 by splitting the middle term.
Factorization by Inspection
Example 15
factorize x2 + 3x + 2 by inspection.
Example 16
factorize 4x2 + 5x – 6 by inspection.
Exercise 1
Factorization Exercise;
