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LINEAR EQUATIONS

This is one of the most powerful concept of Algebra and students are seriously reminded to master this chapter. Those who can master this chapter will find it quite easy to cope with Additional Mathematics or any other calculation subject in the near future.

What makes this chapter interesting is that, the method of solution comes in various form and manner. Today we will be looking into 2 well known methods in solving question that is related to linear equation.

Example : Finding the value of x and y using 2 given equations of 3x + 2y = 13 and x + 4y = 11

Method 1 is by substituting one of the unknown value from one of the given equation into the other equation.

Let 3x + 2y = 13 be (I) and x + 4y = 11 be (II)

From (II); x = 11 – 4y and now known as (III)

By substituting (III) into (I)…take note that (III) is generated from (II) and such is proper to apply substitution  in (I)

3(11 – 4y) + 2y = 13

33 – 12y + 2y = 13

33 – 10y = 13

10y = 20

as such, y = 2

In order to find the value of x, now we substitute y = 2 into any chosen equations of which in this case, we substitute into (III)

 x = 11 – 4(2)

x = 11 – 8 = 3

Method 2 is by neutralizing one of the unknown by the way of addition or subtraction.

Let 3x + 2y = 13 be (I) and x + 4y = 11 be (II)

Say we wish to neutralize the x value, to do that we have to multiply (II) with 3

By doing so, we have 3(x + 4y = 11) of which will gives 3x + 12y = 33, now known as (III)

Subtract (III) with (I):

3x + 12y = 33 – (3x + 2y = 13)

3x – 3x + 12y – 2y = 33 – 13

10y = 20

y = 2

Later we substitute the value of y = 2 into (II) which will gives x + 4(2) = 11

Thus will give a value of x = 3