SOLVING EQUATIONS BY SUBSTITUTION METHOD

Solve the following systems of equations by substitution.

Example 1 : 

x - 5y + 17  =  0

2x + y + 1  =  0

Solution :

x - 5y + 17  =  0 -----(1)

2x + y + 1  =  0 -----(2)

Step 1 :

Solve (1) for x. 

x - 5y + 17  =  0

Subtract 17 from each side. 

x - 5y  =  -17

Add 5y to each side.

x  =  5y - 17 -----(3)

Step 2 : 

Substitute (5y - 17) for x in (2). 

(2)-----> 2(5y - 17) + y + 1  =  0

10y - 34 + y + 1  =  0

11y - 33  =  0

Add 33 to each side.

11y  =  33

Divide each side by 11.

y  =  3

Step 3 :

Substitute 3 for y in (3).

(3)-----> x  =  5(3) - 17

x  =  15 - 17

x  =  -2

Therefore, the solution is 

(x, y)  =  (-2, 3)

Example 2 : 

5x + 3y - 8  =  0

2x - 3y + 1  =  0

Solution :

5x + 3y - 8  =  0 -----(1)

2x - 3y + 1  =  0 -----(2)

Step 1 :

Solve (1) for 3y. 

5x + 3y - 8  =  0

Add 8 to each side. 

5x + 3y  =  8

Subtract 5x from each side.

3y  =  8 - 5x -----(3)

Step 2 : 

Substitute (8 - 5x) for 3y in (2). 

(2)-----> 2x - (8 - 5x) + 1  =  0

2x - 8 + 5x + 1  =  0

7x - 7  =  0

Add 7 to each side.

7x  =  7

Divide each side by 7.

x  =  1

Step 3 :

Substitute 1 for x in (3).

(3)-----> 3y  =  8 - 5(1)

3y  =  8 - 5

3y  =  3

Divide each side by 3.

y  =  1

Therefore, the solution is 

(x, y)  =  (1, 1)

Example 3 : 

4x - 7y  =  0

8x - y - 26  =  0

Solution :

4x - 7y  =  0 -----(1)

8x - y - 26  =  0 -----(2)

Step 1 :

Solve (1) for 4x. 

4x - 7y  =  0

Add 7y to each side. 

4x  =  7y -----(3)

Step 2 : 

Substitute 7y for 4x in (2). 

(2)-----> 8x - y - 26  =  0

2(4x) - y - 26  =  0

2(7y) - y - 26  =  0

Simplify.

14y - y - 26  =  0

13y - 26  =  0

Add 26 to each side.

13y  =  26

Divide each side by 13.

y  =  2

Step 3 :

Substitute 2 for y in (3).

(3)-----> 4x  =  7(2)

4x  =  14

Divide each side by 4.

x  =  3.5

Therefore, the solution is 

(x, y)  =  (3.5, 2)

Example 4 :

Your family opens a bed-and-breakfast. They spend $600 preparing a bedroom to rent. The cost to your family for food and utilities is $15 per night. They charge $75 per night to rent the bedroom

a)  Write an equation that represents the costs.

b) Write an equation that represents the revenue 

c)  Write the system of linear equations for this problem.

Solution :

The amount spend for preparing bedroom

Cost of food per night = $15

Let x be the number of nights to be rented.

a) Cost = 600 + 15x

b) To rent the bedroom, they charge $75 per night.

Revenue = 75 x

c) C = 600 + 15x and R = 75x these are the system of linear equations representing the situation.

Example 4 :

You sell small and large candles at a craft fair. You collect $144 selling a total of 28 candles. How many of each type of candle did you sell?

Solution :

Let x be the number of small candles.

Cost of each small candle = $4

Let y be the number of large candle.

Cost of large candle = $6

Total amount = $144

4x + 6y = 144 -----(1)

Total number of candles sold = 28

x + y = 28 -----(2)

x = 28 - y

Applying the value of x in (1), we get

4(28 - y) + 6y = 144

112 - 4y + 6y = 144

112 + 2y = 144

2y = 144 - 112

2y = 32

y = 32/2

y = 16

When y = 16

x = 28 - 16

x = 12

So, number of small candles is 12 and number of large candles is 16.

Example 5 :

A drama club earns $1040 from a production. A total of 64 adult tickets and 132 student tickets are sold. An adult ticket costs twice as much as a student ticket. Write a system of linear equations that represents this situation. What is the price of each type of ticket?

Solution :

Let x be the cost of student ticket and y be the cost of adult ticket.

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

Total number of adult tickets sold = 64

Number of students ticket sold = 132

132x + 64y = 1040 ------------(2)

Applying (1) in (2), we get

132x + 64(2x) = 1040

132x + 128x = 1040

260x = 1040

x = 1040/260

x = 4

Applying the value of x, we get

y = 2(4)

= 8

So, cost of each student ticket is $4 and cost of each adult ticket is $8.

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