# FREE Basic Math Concept-04

FREE Basic Math Concept-04

## Algebra:

### Equations:

Any expression that includes an equality sign ‘ = ’ and either a variable x, ‘x2, y etc’ or a constant value ‘2, 3 or 4 etc.’ is known as equation. For instance,

a = 4,   x + y = 5,   2x + y = 15   and   x2 – 10x + 24 = 0   etc,   are all equations.

There are two types of equations:

• Simple equations

### Simple Equations:

Those equations, in which the maximum value of exponent of a variable ‘x or y’ is not greater than 1 are simple equations. e.g.,

x + y = 5

and

2x + y = 15

These are simple equations because the maximum value of the exponent of variable is 1 for both ‘x’ and ‘y’.

#### Solving Equations Simultaneously:

Let’s consider the above two equations and solve it for the values of the variables ‘x’ and ‘y’.

x + y = 5 —————————– (Eq. 1)
2x + y = 15 ————————– (Eq. 2)

Now, as there are two same kind of variables exist in both of the equations, so in order to find the one variable (let’s suppose ‘x’) we need to eliminate the other variable (i.e y).

For that purpose, subtract first equation (Eq. 1) from second equation (Eq. 2), i.e., So, we got,

x = 10

Now, put this value of ‘x’ in any of the equations above to find the value of ‘y’

Let’s put value of ‘x’ in first equation (Eq. 1), we’ll get,

(10) + y = 5

By subtracting both side of the equation by 10, we’ll get:

(10 + y) – 10 = 5 – 10

10 + y – 10 = – 5

So, y = – 5 Answer

#### Another Example:

If the two equations were,

2x – 3y = 10, and 3x + 2y = 80 then solve it for value of x and y?

2x – 3y = 10 ——————————— (Eq. 1)
3x + 2y = 80 ——————————— (Eq. 2)

Again, we need to eliminate one variable (let’s say ‘y’) to find the value of other variable (x). For that purpose, we need make the coefficients of y to be same in both of the equations, irrespective of its sign (either – or +). In first equation, the coefficient of y is ‘–3’ while in second equation, the coefficient of y is ‘2’. If we multiply coefficient of y in second equation (i.e. 2) with the coefficient of y in first equation, then the coefficient of y in first equation would become ‘–6’ while if we multiply the coefficient of first equation ‘3’ (irrespective of ‘–’ sign) with the coefficient of the second equation ‘2’, we’ll get the coefficient of second equation to be 6. But remember, while multiplying, we have to remember the rules of multiplication, i.e., if we multiply any equation with a number, we need to multiply it with the whole equation on both sides rather than to only one coefficient or value. i.e.,

By multiplying (Eq. 1) with 2 on both sides, we’ll get,

2 (2x – 3y) = 2 (10)

4x – 6y = 20 ———————————- (Eq. 3)

Similarly,

By multiplying (Eq. 2) with 3 on both sides (to make the y-coefficient same in both equations), we’ll get,

3(3x + 2y) = 3(80)

9x + 6y = 240 ———————————- (Eq. 4)

##### Simultaneous process:

Now, you can see that the coefficients of y in both equations (i.e (Eq. 3 and Eq. 4)are same but reverse in signs. So If we add the two equations (Eq. 3) and (Eq. 4), we can eliminate variable y, to find value of variable x, i.e.,

4x – 6y = 20
9x + 6y = 240 13x = 260

Now, dividing both side by 13,

13x13

x = 26013

So, we got,

x = 20

Now, put this value of ‘x’ in any of the equations (Eq. 1, Eq. 2, Eq. 3 or Eq. 4) above to find the value of ‘y’

Let’s put value of ‘x’ in first equation (Eq. 1), we’ll get.

2(20) – 3y = 10

40 – 3y = 10

Adding both side of the equation by 3y – 10, we’ll get:

(40 – 3y) + (3y – 10) = 10 + (3y – 10)

⇒ 40 – 3y + 3y – 10 = 10 + 3y – 10

⇒ 40 – 10 = 3y

⇒ 30 = 3y

Now, dividing both sides by 3, we’ll get,

Those equations, which value of exponent of a variable ‘x or y’ always 2, are quadratic equations. For instance,

x2 = 1,   and   x2 – 10x + 24 = 0, are quadratic equations.

The general form of quadratic equation is:

ax2 + bx + c = 0

Where,

a = Coefficient of x2

b = Coefficient of x

c = Constant (i.e an integer)

There are two ways to solve such equations:

• By Factorization

##### By Factorization:

This way is mostly useful when the coefficient of x2 is 1, or can be converted to 1 by taking common etc.

Take a look into the quadratic equation below, and let’s solve it for value of x by using factorizing.

2x2 – 20x = 48

First we need to make it like general form of quadratic equation i.e ax2 + bx + c = 0

⇒ 2x2 – 20x – 48 = 0                          (After subtracting both side by 48)

Now, we need to make the coefficient of x2 to be 1. So let’s take common 2 from left hand side of the equation:

⇒ 2(x2 – 10x – 24) = 0

Now, by dividing this equation by 2 on both sides, we’ll get

x2 – 10x – 24 = 0

Now, we can solve this quadratic equation by factorizing. So let’s factorize the above equation.

Factorizing the quadratic equation involve careful breakdown of the coefficient of x such that the sum of the breakdown values must be equals to the coefficient of x including the sign (i.e whether + or –). Furthermore, the product of the breakdown values must be equals to the value constant (i.e c in general form of equation).

So let’s factorize the quadratic equation: x2 – 10x – 24 = 0

x2 + (2x – 12x) – 24 = 0

Note that we did not breakdown –10 into –6 and –4. The reason is that, although the sum of these two gives back –10, but the product of these two does not give the constant –24. The product of –6 and –4 gives +24. Therefore, we cannot breakdown –10 into –6 and –4. And hence, we have breakdown –10 into +2 and –12 so that the sum must give us back –10, while the product must give the constant –24.

Now, we have

x2 + (2x – 12x) – 24 = 0

x2 + 2x – 12x – 24 = 0

⇒ (x2 + 2x) – (12x + 24) = 0

It is very important to note that by placing parenthesis, signs inside the parenthesis will be changed, because parenthesis has a negative sign on its left. Many students commit such silly mistakes and thereby not able to succeed in exams. So be careful from onward 🙂

To avoid such mistake, it’s better to do following steps at beginner level,

###### Factorization steps:

x2 – 10x – 24 = 0

x2 + 2x – 12x – 24 = 0

⇒ (x2 + 2x) + (–12x – 24) = 0

By taking –ve sign common (basically, –ve sign = –1)

⇒ (x2 + 2x) – (12x + 24) = 0

These are the steps that I had skipped before. Now,

x(x + 2) – 12(x + 2) = 0

Now, by taking (x + 2) common,

⇒ (x + 2) {x – 12} = 0

⇒ (x + 2) (x – 12) = 0 —————- (eq. 5)

Now, remember that when that product of two values equals to 0, then at least one of the two values must be zero. i.e

Either,

(x + 2) = 0               or               (x – 12) = 0

If the general form of quadratic equation is considered, the quadratic formula is: Where the general form of quadratic equation, which we have discussed before, is:

ax2 + bx + c = 0

Let’s solve the same quadratic equation as we already solved using factorizing before.

x2 – 10x – 24 = 0

First, make this equation similar to the form of a general form of quadratic equation i.e

x2 + (–10)x + (–24) = 0

Now, let’s compare the coefficients of this equation with the general form of quadratic equation,

ax2 + bx + c = 0

a = 1       b = –10       c = –24  ## Other important Algebraic rules:

Any number raised to power 0 is always equals to 1. Even 0 raised to power 0 is also equals to 1. i.e:

x0 = 1

And

00 = 1

But only few mathematicians suggest that 00 = undefined while majority suggest to consider it as 1. So let’s not involve ourselves in such dispute, and consider it simply 1. As this expression is yet disputed, so it will never be tested in exams. HURRY!

Also, remember that when two or more exponential expressions are adding or subtracting, try to take common to simplify it, i.e This concept is mostly tested in Algebra.

Now, you are ready to attempt a mini Quiz. Click on ‘Mini Quiz’ button below:

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Notify of Inline Feedbacks Alina Hayder
September 10, 2017 11:59 pm

0^0 = undefined Tayyab Aslam
April 5, 2018 3:03 pm

Zero exponent zero is undefined. Please correct the mistake above. Thank you Tayyab Aslam
April 6, 2018 2:00 pm

First thing, Wikipedia is not an authentic reference. Second thing, it is not disputed. It is unambiguous. Let me explain this: 0^1, 0^2 is going to be zero. Because we know, anything with exponent 1 means it occurs once. For example, 1^1 = 1, 1^2 = 1*1 =1, and so one 0^1 means zero occurs for one. Or simply, 0^1 means 0; every entity has an exponent 1, it is understood. No need to write the exponent. But 0^0, it is undefined. How many times are you going to write 0?? For zero times?? Can you count zero?? No, we… Read more » Tayyab Aslam
April 5, 2018 3:02 pm 