Algebra
ClementRadcliffe, Contributor

Students take notes at Bridgeport High School. - Anthony Minott/Freelance Photographer

Let us begin today's lesson by reviewing the answers to last week's homework.

1. State each of the following numbers correct to the number of decimal places given in brackets:

(a) 5.08 (1 d.p.)
(b) 289.438 (2 d.p.)
(c) 0.0088 (3 d.p.)

SOLUTION

(a) 5.1
(b) 289.44
(c) 0.009

2. Write each of the following numbers correct to the number of significant figures indicated in each bracket:

(a) 35.93106 (2 s.f.) = 36
(b) 25.37 (1 s.f.) = 20

N.B. The answer is not 2, as 2 is NOT approximately equal to 25.37

(c) 37.8567 (3 s.f.) = 37.9

3. Write in standard form:

(a) 0.005941 = 5.94 x 10-3

(b) 79,376.82 = 7.94 x 104

4. Expand the following:

(a) (M + 3) (M - 3) = M² + 3M - 3M - 9 = M² - 9

(b) (K - 3) (K + 5) = K² - 3K + 5K - 15 = K² + 2K - 15

We will now turn to ALGEBRA by reviewing ALGEBRAIC FRACTIONS.

ALGEBRAIC FRACTIONS

The method of simplifying algebraic fractions is the same as that is used for vulgar fractions. This is also true for addition or subtraction of algebraic fractions. It follows then that you must know the method used to find L.C.M.

Example 1

The L.C.M. of the denominators is 10

(I am sure that you recall that the negative sign in front of the brackets will change the sign within the bracket)

Example 2

The L.C.M. of the denominators is p(2p - 3).

Now attempt the following

Simplify

LINEAR EQUATIONS

The inclusion of the EQUAL sign differentiates an EQUATION from an algebraic expression. This point is commonly missed by students who sometimes attempt to solve algebraic expressions. Do not fall into this trap.

The following points should be noted:

• Equations identify either the relationship between variables or the value of a variable.
  
  
• The value of the variable is maintained by performing identical operations on both sides of the equation.
  
  
• The methods of clearing brackets and simplifying algebraic expressions are usually required to find solution of equations.
  
  
• In order to solve the equation, one approach is to simplify each side of the equation and then equate both sides.

Example

June 1996, No. 2 (d)

Considering the left-hand side, the L.C.M. of 3 and 4 is 12.

Equating both sides

ALTERNATIVELY, you may multiply all terms by the L.C.M. of the denominators.

Multiply both sides by 12:

ie. 3(4x + 5 - 4(9 + 2x) = 0

ie. 12x + 15 - 36 + 8x = 0

ie. 4x - 21 = 0

You are encouraged to select and practise the method with which you are more comfortable.

Please attempt the following for homework.

Remember, you must practise as many exercises as possible if you are to succeed. I urge you to find examples in your textbooks and work on them.

Clement Radcliffe is the principal of Glenmuir High School in May Pen.