Any major competitive exam in India, such as SBI PO bank exam, IBPS bank exam, SSC CGL 2017, and more are known to turn up the difficulty level of the exam with every passing year. An aspirant looking to join any of the government jobs faces his biggest challenge in the Quantitative Aptitude section of these exams.

Generally speaking, the questions in this section would have a situation explained where few conditions are provided. The trick is to understand the given conditions correctly and get the required answer for the question, all under the matter of a few seconds. All these conditions can be framed in terms of equations and to get the required answer, the aspirant needs to solve the equation correctly.

It is very important to understand how to frame the different equations and how to solve the equation to get the required answer.

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**What Are The Different Types of Equations**

1.Linear Equation

2.Quadratic Equation

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**What is Linear Equation?**

In a Linear Equation, the power or the degree of the variable(s) is 1 and it does not occur in any other powers.

Explanation: Variable is the unknown value that needs to be determined. And constants are the values already given in the conditions in the question.

In the general form given above,

ax + b = 0 This is a linear equation in 1 variable, which is x

ax + by = c This is a linear equation in 2 variables, i.e., x and y

ax + by + cz = d This is a linear equation in 3 variables, i.e., x, y and z.

As in equations shown above, the highest power of the variables is one.

Points to remember: In a linear equation with 1 variable, only one such equation is needed to find the solution. However, in a case of equations with 2 variables, where we are required to find values of 2 unknown, we need 2 linear equations. And quite similarly, to solve and find the values of 3 variables in a linear equation, we need to have 3 different linear equations.

In bank exams and many other competitive exams, linear equations with 3 variables are very rare.

Example 1:

Linear equation with 1 variable:

4x – 12 = 0

Solution:In order to solve a linear equation, we must take the known values on the right-hand side by changing the signs.

4x – 12 = 0

4x = 12

x = 12/4 = 3

**x = 3**

4 * 3 – 12 = 0

12 – 12 = 0

Linear equations with 2 variables:

3x + 2y = 13

5x + 3y = 21

Solution:To solve linear equations with 2 variables, we must eliminate either one of them, i.e., x or y. And in order to do that, the coefficients of the variables in the 2 equations must be equated.

To equate the coefficients of x in both these equations, we must multiply the first equation with the coefficient of the second and the second equation with that of the first.

First Equation:

(3x + 2y = 13) * 5

15x + 10y = 65

Second Equation:
(5x + 3y = 21) * 3

15x + 9y = 63

15x + 10y = 65

_15x + 9y = 63

**y = 2**

Since we now have y, we can substitute the variable with its value in any of the equation and find out the value of the other variable, x.

3x + 2y = 13

3x + 2(2) = 13

3x + 4 = 13

3x = 9

**x = 3**

Solve the equations given below and establish the relationship between x and y.

6x + 7y = 93

3x + 2y = 33

Solution:We can easily equate the coefficients of x in the 2 equations by multiplying the second equation with 2.

(3x + 2y = 33) * 2

6x + 4y = 66

To get the value of x, we must subtract the 2 equations:
6x + 7y = 93

_6x + 4y = 66

3y = 27

**y = 9**

6x + 7y = 93

6x + 7*9 = 93

6x + 63 = 93

6x = 93 – 63

6x = 30

**x = 5**

Since y = 9 and x = 5, the relationship between x and y is said to be

**x<y**
Example 4:

Solve the equations and establish the relationship between the variables x and y.

√36 x + √64 = 0

√81 y + 4² = 0

Solution:

First Equation:

√36 x + √64 = 0

6x + 8 = 0

x = -8/6

**x = -4/3**

Second Equation:

√81 y + 4² = 0

9y + 16 = 0

**y = -16/9**

Comparision of fractions:

To establish the relationship between x and y, simply cross multiply the two variables.

x * y

- 4/3 * - 16/9

- 36 > - 48

**x>y**

If not for the minus sign, then 36 < 48 which would give you the incorrect solution.

Points to remember:

1.You must remember to not discard or cancel out the minus sign, which would eventually give you the relationship between x and y.

2.If we are required to compare fractions, cross multiply them.

Example 5:

Solve the equations and establish the relationship between the variables x and y.

9 ⁄√x + 19 ⁄√x = √x

y⁵ - [(2 * 14)

^{11/2 }]/√y = 0
Solution:

First Equation:

9 ⁄√x + 19 ⁄√x = √x

Addtion of fractions:

To add two fractions, we must calculate the LCM of the denominators of the two fractions.

(9 + 19)/√x = √x

28 = √x * √x

**x = 28**

Second Equation:

y⁵ - [(2 * 14)

^{11/2 }]/√y = 0
y⁵ = [28

^{11/2 }]/√y
y⁵ * √y = 28

^{11/2}
y 28

^{(5 + ½) }=^{11/2}
y 28

^{(11/2) }=^{11/2}
Since the powers are equal on either side of the equation, the bases are also equal.

**y = 28**

Since x = 28 and y = 28, the relationship between x and y is said to be

**x=y**

Example 6: The cost of 4 chairs and 1 table is Rs. 3400. The cost of 5 chairs and 3 tables is Rs. 6000. What is the cost of one chair and one table?

Solution: Let us consider the chair to be variable x and table to be variable y.

First Equation: 4x + y = 3400

Second Equation: 5x + 3y = 6000

First Equation:

(4x + y = 3400) * 3

12x + 3y = 10200

By subtracting the second equation from the first:

12x + 3y = 10200

_5x + 3y = 6000

7x = 4200

**x = 600**

Substituting the value of x, in any of the given equations we would get the value of y.

4x + y = 3400

4 * 600 + y = 3400

y = 3400 – 2400

**y = 1000**

Hence, cost of one chair is Rs. 600 and the cost of one table is Rs.1000.

To know about what a Quadratic Equation is and How to solve them easily, watch out our next blog. Do let us know how this blog has helped you understand what is linear equation and how to solve them.

Keep practising. Because there is no other way to master these techniques.

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