SOLVED DIFFERENT PROBLEMS USING ELIMINATION METHOD

Question 1 :

Solve the pair of linear equations using elimination method.

10/(x + y) + 2/(x –y)  =  4 and 15/(x + y) – 5/(x – y)  =  -2

Solution :

Let 1/(x + y) = a  and 1/(x – y) = b

10a + 2b = 4  --------(1)

15a – 5b = -2  --------(2)

(1) ⋅ 5 + (2) ⋅ 2 

50a + 10b + (30a - 10b) = 20 - 4   

80 a  =  16

a = 16/80

a = 1/5

Substitute a = 1/5 in the first equation

10(1/5) + 2b  =  4

2 + 2b  =  4

2b  =  2

b  =  1

(3) + (4)

x + y + x - y = 5 + 1

2x  =  6

x  =  3

By applying the value of x in (3), we get 

3 + y  =  5

y  =  2

Question 2 :

Solve the pair of linear equations using elimination method.

1/(3x + y) + 1/(3x – y)  =  3/4

and

 1/2(3x + y) - 1/2(3x – y)  =  -1/8

Solution :

1/(3x + y) = a and  1/(3x – y) = b

a + b = 3/4

4(a + b) = 3

4a + 4b = 3 ------(1)

(a/2) – (b/2)  =  -1/8

4a – 4b  =  -1------(2)

(1) + (2)

4a + 4b + 4a - 4b = 3 - 1

8a = 4  ==> a = 1/2

By applying the value of a in (1), we get 

4(1/2) + 4b  =  3

2 + 4b  =  3

4 b  =  1

b  =  1/4

(3) + (4)

3 x + y + 3x - y = 2 + 4

6x = 6

x = 1

By applying x = 1 in the (3) equation, we get

3(1) + y = 2

y = 2 – 3

y = -1

Question 3 :

In a football game, all of the home team’s points are from 7-point touchdowns and 3-point field goals. The team scores six times. Write and solve a system of linear equations to find the numbers of touchdowns and field goals that the home team scores.

Solution :

Let t be the number of touchdown and f be the field goals.

The system of equations representing the situation.

7t + 3t = 26 --------(1)

t + f = 6 --------(2)

f = 6 - t

7t + 3(6 - t) = 26

7t + 18 - 3t = 26

4t = 26 - 18

4t = 8

t = 8/4

t = 2

f = 6 - 2

f = 4

Number of touchdown goals = 2 and number of field goals is 4.

Question 4 :

You are an ecologist studying the populations of two types of fish in a lake. Use the information in the table to predict when the populations of the two types of fish will be equal.

Solution :

Let x be the number of years.

-250x + 5750 = 400x + 2825

Solving for x, we get

-250x - 400x = 2825 - 5750

-650x = -2925

x = 2925/650

x = 4.5

So, the populations of the two types of f i sh will be equal in 4.5 years.

Question 5 :

Your family needs to rent a car for a week while on vacation.

• Company A charges $3.25 per mile plus a flat fee of $125 per week.
• Company B charges $3 per mile plus a flat fee of $150 per week.

After how many miles of travel are the total costs the same at both companies?

Solution :

Let x be the number of miles traveled.

3.25x + 125 = 3x + 150

3.25x - 3x = 150 - 125

0.25x = 25

x = 25/0.25

x = 100

So, the required number of miles traveled is 100.

Question 6 :

You need to hire a catering company to serve meals to guests at a wedding reception. Company A charges $500 plus $20 per guest. Company B charges $800 plus $16 per guest. For how many guests are the total costs the same at both companies?

Solution :

Let x be the number of guests.

Charges from Company A : 500 + 20x

Charges from Company B : 800 + 16x

500 + 20x = 800 + 16x

500 - 800 = 16x - 20x

300 = 4x

x = 300/4

x = 75

So, the number of guests is 75.

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