Algebraic fraction
Clement Radcliffe, Contributor

Let us begin this week'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) 5.05 (1 d.p.) = 5.1
(b) 286.598 (2 d.p.) = 286.60
(c) 0.0088 (3 d.p.) = 0.009

NB: In (b) I do hope that you realise that 286.6 would be incorrect.

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

(a) 46.93106 (2 s.f.) = 47
(b) 45.37 (1 s.f.) = 50
(c) 37.8567 (3 s.f.) = 37.9

3. Write in standard form:

(a) 0.003921 = 3.92 x 10-3 (The number is multiplied by 1000).

(b) 49,376.82 = 4.94 x 104

Please be reminded that you may check your answer by converting the standard form to a decimal fraction.

4. Expand the following:

(a) (M + 2) (M - 2)
(b) (K - 4) (K + 1)

(a) (M + 2) (M - 2) = M2 + 2M - 2M - 4 = M2 - 4
(b) (K - 4) (K + 1) = K2 - 4K + K - 4 = K2 - 3K -4

We now return to algebra by reviewing algebraic fractions.

ALGEBRAIC FRACTIONS

The method of simplifying algebraic fractions is the same as that 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

(a) 2(5x - y) - (3x - y)

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 in 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

Here is another example:

June 1995, No. 2 (b)

Using an alternate method, multiply both sides by 6 (the L.C.M of 2 and 3)

ie. 3(P - 1) - 2(P - 2) = 6
ie. 3P - 3 - 2P + 4 = 6
ie. P + 1 = 6
ie. P = 5

You may substitute P = 5 in the original equation to test the answer as follows:

Considering the left hand side, if P = 5,

ie. The equation is satisfied.

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

Please attempt the following for homework.

1. 2(x + 2y) - 4(x - y) =

(A) - 2x + 8y
(B) - 2x
(C) 6x + 8y
(D) - 2x + y

2. Solve (5P - 2) - 3(P + 4) = 8

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 them.

Clement Radcliffe teaches at Glenmuir High School.