Algebra Part 11 - An Application

Apply what you have learnt in this series of lessons and solve the following problem:

A person drives from Town A to Town B, a distance of 240 km.

While  returning from town B, due to heavy traffic, his speed was reduced by 20 km/h and the return journey took 1 hour more than his onward journey.

Form a quadratic equation and find his onward speed from Town A to town B. 

Onward Journey from Town A to Town B

Let his speed = x km/h.

Time = distance/speed

= 240/x hours.

Return Journey from Town B to Town A

His speed was reduced by 20km/h due to heavy traffic.

So speed = (x – 20) km/h.

Time = 240/(x – 20) hours.

Solve to find x:

(x – 80) (x + 60) = 0

(x – 80)= 0 or (x + 60) = 0

x = 80 or x = – 60 (rejected)

His onward speed from Town A to town B is 80 km/h.

Algebra - Part 10 - Solving Quadratic Equation by Using Formula

Algebra - Part 9 - Solve Quadratic Equation by Factorisation

Algebra - Part 8 - Quadratic Equation, Discriminant, Nature of Roots

Linear equation in x

x = 3

The highest power of x in the above equation is 1. 
So it is called Linear Equation.

Quadratic equation in x

x2 + 2x + 1 = 0

In the above equation, the highest power of x is 2. 
So this equation is called Quadratic Equation.

Some other quadratic equations

x2  – 25 = 0          x2 + 5x = 0

General form of a Quadratic Equation
ax2 + bx + c = 0 where a,b,c are constants and a ¹  0.

A quadratic equation has two solutions, known as roots.

Roots of a quadratic equation may be equal or distinct (not the same value) . 
Some quadratic equations do not have real roots.

Nature of roots can be determined by the value of the discriminant, D =  b2 – 4ac.

If D > 0, the roots are distinct.

If D = 0, the roots are equal.

If D < 0, then there are no real roots.

For, x2 + 2x + 1 = 0, 

a = 1, b = 2, c = 1

D =  b2 – 4ac

D =  22 – 4(1)(1) = 0

So this equation has equal roots.

For, 2x2 + 4x – 5 = 0, 

a = 2, b = 4, c =  – 5

D =  b2 – 4ac

D =  42 – 4(2)(– 5) = 56 > 0

So the roots of this equation are distinct.

Algebra - Part 7 - Elimination Method

Elimination Method

2a + b = 13 ……..(1)

3a + 2b = 21…… (2)

Decide which variable you want to eliminate.

Let us eliminate the variable  ‘b’.

Look at the number before ‘b’.

1 and 2.

We have to make these two numbers equal.

Let us multiply (1) by 2.

(1) x 2     4a + 2b = 26…….(3)

(2)          3a + 2b = 21…… (2)

Now we made the numbers equal. (2, 2)

Look at the sign of the two numbers.  (+2 , +2)

If they are of the same sign (+,+) or (– , – ) to eliminate ‘b’, we need to subtract the equations.

If they are of different sign, (+, – ), we need to add the equations.

Since both numbers are of the same sign, we subtract.

4a + 2b = 26 ……(3)

3a + 2b = 21 ……(2)

(3) – (2),

(4a + 2b) – (3a + 2b) = 26 – 21

4a + 2b – 3a – 2b = 26 – 21

                            a = 5

Substitute a = 5 in (1),

(1)  2a + b = 13

          2(5) + b = 13

            10 + b = 13

                     b = 13 – 10

                     b = 3

 

Algebra - Part 6 - Substitution Method

Equation

a = 3

a has a value of 3

If a + b = 8, what are the values of a and b?

a = 3, b = 5, Check: 3 + 5 = 8

a = 7, b = 1, Check: 7 + 1 = 8

and a lot more answers.

There is no unique solution, because we have two unknowns (a,b) but only one equation (a + b = 8)

Simultaneous Equations

2a + b = 13

3a + 2b = 21

These two equations are called simultaneous equations.

How to find the solution of simultaneous equations?

There are many ways to solve simultaneous equations.  
In this lesson, we will solve by substitution method.

Substitution Method

2a + b = 13 ……..(1)

3a + 2b = 21…… (2)

from (1) we get,

b = 13 – 2a ….(3)

Substituting (3) into (2),

3a + 2b = 21 …..(2)

3a + 2(13 – 2a) = 21

3a + 26 – 4a = 21

26 – a = 21

      – a = 21 – 26 = – 5

         a = 5

b = 13 – 2a 
b = 13 – 2(5)

b = 3 

Algebra - Part 5 - Make as Subject

Make ‘a’ as the subject:

5a + 3b = 3a + 8b + 5

That means keep only the variable ‘a’ on one side of the equation. All other variables and constants must be moved to the other side.

5a + 3b = 3a + 8b + 5

5a + 3b – 3a = 8b + 5

2a + 3b = 8b + 5

2a = 8b + 5 – 3b

2a = 5b + 5

a = (5b + 5) / 2

While moving variables/constants, start with those furthest from variable ‘a’ than nearer to ‘a’.

Algebra - Part 4 - Solving Equations

Find the value of a:
a + 3 = 5

We only need the variable a on the left side of this equation.

Hence we subtract 3 from both sides.

To maintain the equality, always do the same operation for both sides of the equation.

a + 3 – 3 = 5 – 3

             a = 2

Alternatively, you can move +3 from left side to right side. When you move +3 to right side, it becomes – 3.

a + 3 = 5

      a = 5 – 3

      a = 2

When you move a constant or variable to the other side of the equation:

+ becomes –

– becomes +

x becomes ¸

¸ becomes x

Examples

b – 6 = 10

b = 10 + 6

b = 16

5c = 15

c = 15 ¸  5

c = 3

d/2  = 6
d = 6 x 2

d = 12

Algebra - Part 3 - Expansion and Factorisation

During addition and multiplication, you can switch positions without affecting the answer.

a + b = b + a      

ab = ba

During subtraction and division, if you switch positions, the answer will change.

a – b ¹ b – a   

a ¸ b ¹ b ¸ a

To expand, multiply each and every term inside the brackets.

4(3x – 5)

= 4 (3x) +  4 (– 5)

= 12x – 20

(2x) (3x – 5)

= (2x) (3x) +  (2x) (– 5)

= 6x2 – 10x

(2x + 3y) (3x – 5)

= (2x) (3x – 5) + (3y) (3x – 5)

= (2x) (3x) +  (2x) (– 5)
+ (3y) (3x) +  (3y) (– 5)

= 6x2 – 10x + 9xy  – 15y

To factorise, take out common terms.

2a2b2 + 10a2b + 20b

2 and b are common in all these terms.

2a2b2 + 10a2b + 20b  = 2b (a2b + 5a2 + 10)

Factorise:

2ax + 3ay + 2bx + 3by

First two terms: a is common 

Last two terms: b is common.

= a (2x + 3y) + b (2x + 3y)

Now (2x + 3y) is common

= (2x + 3y) (a + b)

Algebra - Part 2 - Constant, Variable, Term

In Algebra,

5 is called a Constant.

x is called a  Variable.

5x is called a Term.

Like Terms

5x   8x   – 4x

All have same variable, x

5xy   8xy   – 4xy

All have same variables, xy

Unlike Terms

5x   8y   – 4z

Not all have same variable

5xy   8yz   – 4xz

Not all have same variables

Only like terms can be added or subtracted.

5x  +  8x = 13x

5x –  8x = – 3x

5ab + 3ab = 8ab

25xy – 10xy = 15xy

3x + 2x + 4y – 6y = 5x – 2y

To multiply or divide, terms need not be like terms.

(x) (y) = xy      

(ab) (c) = abc

(2x) (3y) = 6xy

5ab ¸ 5 = ab    

5ab ¸ a = 5b

5ab ¸ ab = 5

Can I multiply or divide like terms too?

Yes, you can.

(x) (x) = x2     

(x) (x) (x) = x3

(2x) (3x) = 6x2   

25ab ¸ 5ab = 5

Algebra - Part 1 - Introduction

Algebra.  Nightmare for many. 

Algebra is used to solve many real life problems.

First, some basics.

Let’s say you, Xavier have $5 in your pocket. You can write this as an EQUATION as follows.

Money = 5 or X = 5.

It simply means X has the value of 5.

Now your elder brother, Yong has $8 MORE than you.

So we can write Y = X + 8.

To find the amount of money he has, put 5 instead of x (we call this as SUBSTITUTE).

5 + 8 = 13.

So your brother has $13

And your younger sister, Zoe has $2 LESS than you, so we can write Z = X – 2 = 5 – 2 = 3

So as an EXPRESSION, what is the total amount of money all three of you have?

It is X + Y + Z

= X + X + 8 + X – 2

= 3X + 6

So 3X + 6 is the total amount of money in TERMS OF X.

But how much is that?

3X + 6

= 3(5) + 6

= 15 + 6

= 21

Now let us check whether this is correct.

You have $5.

Your elder brother has $13

Your younger sister has $3

So the total is 5 + 13 + 3 = $21