CBSE Class 10 Mathematics Chapter 3 - Pairs of linear equations in two variables - Elimination Method

CBSE Class 10 Mathematics  Chapter 3 - Pairs of linear equations in two variables - Elimination Method

Elimination Method:

• The elimination method is known as the addition-subtraction method. It is a technique used in solving systems of linear equations.
• The elimination method is applied when we have a system of linear equations. A system of linear equations consists of two or more linear equations with two or more variables.
• The primary purpose of the elimination method is to find a solution that satisfies all equations in the system simultaneously. This solution corresponds to the values of the variables that make all equations true.
• The elimination method is useful when the coefficients of one of the variables in the equations are opposites (one positive and one negative) or when the coefficients have a common multiple that allows for easy elimination.

Steps in the Elimination Method:

Step 1: Make the Coefficients Equal

• In the first step, we want to make the numbers in front of one of the variables (either x or y) the same in both equations.
• We do this by multiplying both equations by suitable non-zero numbers so that the coefficients (those numbers in front of x or y) become numerically equal.
• Ensure that the coefficients of one of the variables are additive inverses (equal in magnitude but opposite in sign). If they are not, make them additive inverse by multiplying.

Step 2: Add or Subtract to Eliminate a Variable

• Now that we have the coefficients equal, we can add or subtract the two equations to make one of the variables "disappear" (eliminate it).
• The goal is to end up with an equation that has only one variable (either x or y).
• If in Step 2, we obtain a true statement having no variable, then the original pair of equations has infinitely many solutions.
• If in Step 2, we obtain a false statement having no variable, then the original pair of equations has no solution, it is inconsistent.

Step 3: Solve for the Remaining Variable

• After eliminating one variable in the previous step, we now have an equation with just one variable.
• Solve this equation to find the value of that variable.

Step 4: Substitute and Solve

• Take the value you found in Step 3 and substitute it back into one of the original equations.
• This will help you find the value of the other variable.

So, in simpler terms, you make the coefficients the same, add or subtract to get rid of one variable, solve the equation with one variable, and then use that to find the other variable. This method called elimination method.

Example 1: Use elimination method to find all possible solutions of the following pair of linear equations: 2x + 3y = 11 ; 2x — 4y = —24

Solution: 2x + 3y = 11 ………………….. (i)

2x — 4y = —24 …………………… (ii)

In given equation, the coefficients of 2x in both equations are the same and they have same signs.

To make the coefficients of x additive inverses we multiply Equation (i) by -1 (for changing its sign)

(2x + 3y)—1 = 11×—1

—2x—3y = —11 ……………….. (iii)

For eliminating x, add both equation:

(—2x—3y)+(2x — 4y) = (—11) + (—24)

(—2x + 2x) + (—3y—4y) = (—11—24)

—7y = —35

y = 5

Substitute y value from equation (i)

2x + 3(5) = 11

2x + 15 = 11

2x = 11—15

2x = —4

x = —2

So, the value of x is —2 and y is 5.

Example 2: Use elimination method to find all possible solutions of the following pair of linear equations: 2x + 3y = 8 ; 4x + 6y = 7

Solution: 2x + 3y = 8 …………… (i)

4x + 6y = 7 ………………… (ii)

We multiply equation (i) by 2 and equation (ii) by 1 to make them numerically equal.

(2x + 3y) 2 = 8×2

(4x + 6y) 1 = 7×1

We get:

4x + 6y = 16 ……………. (iii)

4x + 6y = 7 ………………. (iv)

Subtract equation (iv) from equation (iii)

(4x + 6y) — (4x + 6y) = (16) — (7)

(4x—4x) + (6y — 6y) = (16 — 7)

0 = 9

This is a false statement. So, pair of equation has no solution.

Example 3: 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?

Solution: Assume Ten’s digit in first number is x and the unit’s digit in first number is y. so first number can be written as: 10x + y

When first number written in reverse x becomes units of digit and y becomes ten’s digit and can be written as: 10y + x

In first condition: The sum of the two numbers is 66, so we can write this equation

(10x + y) + (10y + x) = 66

11x + 11y = 66       (divide both side by 11)

x + y = 6  …………….. (i)

In second condition: The digits of the number differ by 2

x — y = 2 ……………… (ii)

There are two case in this question:

Case 1: If x > y

In this case, Equation 2 becomes:

x – y = 2

System of two equations becomes:

Equation (1): x + y = 6

Equation (2): x – y = 2

Equation (1) and (2) by elimination:

(x + y) + (x —y) = 6 + 2

(x + x) + (y—y) = 8

2x = 8

x = 4

Put x value in equation (1)

4 + y = 6

y = 2

So, the number is 42.

Case 2: If x < y

In this case, Equation 2 becomes:

y – x = 2

System of two equations becomes:

Equation 1: x + y = 6

Equation 2: y – x = 2

Equation (1) and (2) by elimination:

(x + y) + (y —x) = 6 + 2

(x — x) + (y + y) = 8

2y = 8

y = 4

Put x value in equation (1)

x + 4 = 6

x = 2

So, the number is 24.

Verification: 42 + 24 = 66 and 4 – 2 = 2. Also 24 + 42 = 66 and 4 – 2 = 2