### Preliminaries

This section is an extension of the “Linear equations” which most of you might have studied in your high school math class. Therefore, we can start with a discussion of two of the basic algebraic methods for solving a pair of linear equations.

**Algebraic methods of solving a pair of linear equations**

**Substitution method:**

Illustration 1:

**Solve the following pair of equations**

**7****x – ****5****y = ****2**** (****1****)**

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

**Solution:**

**Step ****1****:** We pick either of the equations and write one variable in terms of the other.

It is easier if we first consider the equation in which the coefficient of x or y is 1.

Here in equation (2), the coefficient of x is 1.

Therefore, consider equation (2)

x + 2y = 3

- x = 3 – 2y ——-(3)

**Step ****2**: Substitute the above value of x in equation (1). We get

7(3 – 2y) – 5y = 2

- 21 – 14y – 5y = 2
- 19 = 19y
- y = 1

**Step ****3**: Substituting this value of y in equation (3)

x = 3 – 2 = 1

Thus, x = 1, y = 1 is the solution of the given pair of equations.

**Elimination method:**

Illustration 2:

**The ratio of incomes of two persons is ****9**** : ****7**** and the ratio of their expenditure is ****4**** : ****3****. If each of them manages to save Rs.****2000**** per month, find their monthly incomes.**

**Solution:**

Let us denote the incomes of the two persons by 9x and 7x and their expenditures by 4y and 3y respectively. Then the equations formed in the situation is given by:

9x – 4y = 2000 ————(1)

7x – 3y = 2000 ————(2)

**Step ****1**: Multiply equation (1) by 3 and equation (2) by 4 to make the coefficients of y equal.

27x – 12y = 6000 ———(3)

28x – 12y = 8000 ———(4)

**Step ****2**: Subtract equation (3) from equation (4) to eliminate y, because the coefficients of y are the same.

(28x – 27x) – (0) = 8000 – 6000

- x = 2000

**Step ****3**: Substituting this value of x in equation (1), we get

9(2000) – 4y = 2000

- 4y = 18000 – 2000
- 4y = 16000
- y = 4000

Thus, the solution of the equations is x = 2000, y = 4000.

Therefore, the monthly incomes of the persons are Rs.18,000 and Rs.14,000 respectively.

### Daily life problems

These are the problems which depicts everyday life situations. You should analyse the situation and express it in equations to get the required answer. All these questions demand you to express them in one or two variables, say x and y. A naive outline for approaching these problems is given below.

**Step ****1**: Analyse the question and decide whether to use one variable or two variables (“x” or “x” and “y”)

**Step ****2**: Express the situation in terms of the above variable(s).

**Step ****3**: Solve the equation to find the value of the variable(s).

*Tip: Always assign variable “x” for what you require as the final answer.*

### Age problem

**Age Problem** is one of the most important type in the daily life problems.

In the age problem, statements regarding age of one or more people will be given. You should analyse those statements to find the age of a particular person asked in the question.

*“For detailed theory, refer the book “CSIR-NET General Aptitude – A New Outlook”*