Class 10

### Topic Covered

♦ Elimination Method

### Elimination Method

=> Now let us consider another method of eliminating (i.e., removing) one variable. This is sometimes more convenient than the substitution method. Let us see how this method works.
Q 3139578412

The ratio of incomes of two persons is 9 : 7 and the ratio of their
expenditures is 4 : 3. If each of them manages to save  2000 per month, find their
monthly incomes.
Class 10 Chapter 3 Example 11
Solution:

Let us denote the incomes of the two person 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)

and 7x – 3y = 2000 \ \ \ \  (2)

Step 1 : Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of
y equal. Then we get the equations:

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. So, we get

(28x – 27x) – (12y – 12y) = 8000 – 6000
i.e., x = 2000

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

Remarks :

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.
Q 3169578415

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
Solution:

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.
Q 3179578416

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
Solution:

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.