ELIMINATION METHOD EXAMPLES

Examples 1-5 : Solve the system by elimination.

Example 1 :

x + y = 5

x - y = -1

Solution :

x + y = 5 ----(1)

x - y = -1 ----(2)

(1) + (2) :

Divide both sides by 2.

x = 2

Substitute x = 2 into (1).

2 + y = 5

Subtract 2 from both sides.

y = 3

Therefore, the solution is

(x, y) = (2, 3)

Example 2 :

2x - y = 7

-2x + 3y = -1

Solution :

2x - y = 7 ----(1)

-2x + 3y = -1 ----(2)

(1) + (2) :

Divide both sides by 2.

y = 3

Substitute x = 3 into (1).

2x - 3 = 7

Add 3 to both sides.

2x = 10

Divide both sides by 2.

x = 5

Therefore, the solution is

(x, y) = (5, 3)

Example 3 :

-2x - 9y = -25

-4x - 9y = -23

Solution :

-2x - 9y = -25 ----(1)

-4x - 9y = -23 ----(2)

(1) - (2) :

(-2x - 9y) - (-4x - 9y) = (-25) - (-23)

-2x - 9y + 4x + 9y = -25 + 23

2x = -2

Divide both sides by 2.

x = -1

Subsytitute x = -1 into (1).

-2(-1) - 9y = -25

2 - 9y = -25

Subtract 2 from both sides.

-9y = -27

Divide both sides by -3.

y = 3

Therefore, the solution is

(x, y) = (-1, 3)

Example 4 :

5x + 4y = 9

7x - 6y = 1

Solution :

5x + 4y = 9 ----(1)

7x - 6y = 1 ----(2)

3(1) + 2(2) :

3(5x + 4y) + 2(7x - 6y) = 3(9) + 2(1)

15x + 12y + 14x - 12y = 27 + 2

29x = 29

Divide both sides by 29.

x = 1

Subsytitute x = 1 into (1).

5(1) + 4y = 9

5 + 4y = 9

Subtract 5 from both sides.

4y = 4

Divide both sides by 4.

y = 1

Therefore, the solution is

(x, y) = (1, 1)

Example 5 :

17x + 15y = 79

15x + 17y = 81

Solution :

17x + 15y = 79 ----(1)

15x + 17y = 81 ----(2)

coefficient of x in (1) = coefficient of y in (2)

coefficient of y in (1) = coefficient of x in (2)

(1) + (2) :

32x + 32y = 160

Divide both sides by 32.

x + y = 5 ----(3)

(1) - (2) :

2x – 2y = -2

Divide both sides by 2.

x – y = -1 ----(4)

(3) + (4) :

2x = 4

x = 2

Substitute x = 2 into (3).

2 + y = 5

y = 3

Therefore, the solution is

(x, y) = (2, 3)

Example 6 :

Two numbers add up to 52 and differ by 2. Find the numbers.

Solution :

Let x and y be the twol numbers.

From the information given, we have

x + y = 52 ----(1)

x - y = 2 ----(2)

(1) + (2) :

2x = 54

Divide both sides by 2.

x = 27

Substitute x = 27 into (1). 

27 + y = 52

Subtract 27 from both sides.

y = 25

Therefore, the numbers are 25 and 27.

Example 7 :

In a magic show, the cost of a kid ticket is $3 and that of a adult is $5. In a particular day, there were 325 tickets sold in all for $1275. Find the number of kids tickets and adults tickets sold.

Solution :

Let k and a be the number of kids tickets and adults tickes respectively.

From the information given, we have

k + a = 325 ----(1)

3k + 5a = 1275 ----(2)

5(1) - (2) :

5(k + a) - (3k + 5a) = 5(325) - 1275

5k + 5a - 3k - 5a = 1625 - 1275

2k = 350

Divide both sides by 2.

k = 175

Substitute k = 175 into (1). 

175 + a = 325

Subtract 175 from both sides.

a = 150

Therefore,

number of kids tickets sold = 175

number of adult adults sold = 150

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Integration of tanx

    Mar 18, 24 01:00 PM

    Integration of tanx

    Read More

  2. integration of Sec Cube x

    Mar 18, 24 12:46 PM

    integration of Sec Cube x

    Read More

  3. Solved Problems on Logarithms

    Mar 17, 24 02:10 AM

    Solved Problems on Logarithms

    Read More