Step 3 : Substituting this value of x in (1), we get
`9(2000) – 4y = 2000`
i.e., `y = 4000`
So, the solution of the equations is `x = 2000, y = 4000.` Therefore, the monthly incomes
of the persons are ` 18,000 and ` 14,000, respectively.
Verification : `18000 : 14000 = 9 : 7.` Also, the ratio of their expenditures =
18000 – 2000 : 14000 – 2000 = 16000 : 12000 = 4 : 3
1. The method used in solving the example above is called the elimination method,
because we eliminate one variable first, to get a linear equation in one variable.
In the example above, we eliminated y. We could also have eliminated x. Try
doing it that way.
2. You could also have used the substitution, or graphical method, to solve this
problem. Try doing so, and see which method is more convenient.
Let us now note down these steps in the elimination method :
Step 1 : First multiply both the equations by some suitable non-zero constants to make
the coefficients of one variable (either x or y) numerically equal.
Step 2 : Then add or subtract one equation from the other so that one variable gets
eliminated. If you get an equation in one variable, go to Step 3.
If in Step 2, we obtain a true statement involving no variable, then the original
pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original
pair of equations has no solution, i.e., it is inconsistent.
Step 3 : Solve the equation in one variable (x or y) so obtained to get its value.
Step 4 : Substitute this value of x (or y) in either of the original equations to get the
value of the other variable.
Now to illustrate it, we shall solve few more examples.
Use elimination method to find all possible solutions of the following
pair of linear equations :
2x + 3y = 8 (1)
4x + 6y = 7 (2) Class 10 Chapter 3 Example 12
Multiply Equation (1) by 2 and Equation (2) by 1 to make the
coefficients of x equal. Then we get the equations as :
4x + 6y = 16 (3)
4x + 6y = 7 (4)
Step 2 : Subtracting Equation (4) from Equation (3),
(4x – 4x) + (6y – 6y) = 16 – 7
i.e., 0 = 9, which is a false statement.
Therefore, the pair of equations has no solution.
The sum of a two-digit number and the number obtained by reversing
the digits is 66. If the digits of the number differ by 2, find the number. How many such
numbers are there? Class 10 Chapter 3 Example 13
Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the first number may be written as 10x + y in the expanded form (for example,
56 = 10(5) + 6).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s
digit. This number, in the expanded notation is 10y + x (for example, when 56 is
reversed, we get 65 = 10(6) + 5).
According to the given condition.
(10x + y) + (10y + x) = 66
i.e., 11(x + y) = 66
i.e., x + y = 6 (1)
We are also given that the digits differ by 2, therefore,
either x – y = 2 (2)
or y – x = 2 (3)
If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2.
In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4.
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Verification : Here 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2.