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
• Quadratic 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,

^13x⁄₁₃

x = ²⁶⁰⁄₁₃

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,

y = 10 Answer

Quadratic Equations:

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)

Solving Quadratic Equations:

There are two ways to solve such equations:

• By Factorization
• Using Quadratic Formula

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

⇒ either, x = –2 Answer             or             x = 12 Answer

Using Quadratic Formula:

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

Now, buy putting these values in quadratic formula, we’ll get,

Keep in mind these rules, that will help you later on in your exam preparation.

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.

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