SOLVING SYSTEM OF EQUATIONS

We can use one of the following methods to solve a system of linear equations. 

1. Elimination Method

2. Substitution Method

3. Cross Multiplication Method 

4. Graphing

Example 1 :

Solve by elimination method. 

3x + 4y  =  7

x - 4y  =  -3

Solution :

3x + 4y  =  7 -----(1)

x - 4y  =  -3  -----(2)

In the given two equations, y-term has the same coefficient and different signs. By adding the above two equations, we can eliminate y-term and solve for y. 

(1) + (2) : 

4x  =  4

Divide each side by 4.

x  =  1

Substitute x  =  1 in (1). 

3(1) + 4y  =  7

3 + 4y  =  7 

Subtract 3 from each side. 

4y  =  4

Divide each side by 4.

y  =  1

So, the solution is 

(x, y)  =  (1, 1)

Example 2 :

Solve by elimination method. 

3x + 4y  =  -25

2x - 3y  =  6

Solution :

3x + 4y  =  -25 -----(1)

2x - 3y  =  6  -----(2)

Both x terms and y terms have different coefficients in the above system of equations.

Let's try to make the coefficients of y terms equal.

To make the coefficients of y terms equal, we have to find the least common multiple 4 and 3.

The least common multiple of 4 and 3 is 12.

Multiply the first equation by 3 in order to make the coefficient of y as 12 and multiply the second equation by 4 in order to make the coefficient of y as -12. 

(1) ⋅ 3 ---->  9x + 12y  =  -75

(2) ⋅ 4 ---->  8x - 12y  =  24

Now, we can add the two equations and eliminate y as shown below. 

Divide each side by 17. 

x  =  -3

Substitute -3 for x in (1). 

(1)---->  3(-3) + 4y  =  -25

-9 + 4y  =  -25

Add 9 to each side.

4y  =  -16

Divide each side by 4.

y  =  -4

So, the solution is

(x, y)  =  (-3, -4)

Example 3 : 

Solve for x and y using substitution. 

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 into (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 into (3).

(3)-----> x  =  5(3) - 17

x  =  15 - 17

x  =  -2

So, the solution is 

(x, y)  =  (-2, 3)

Example 4 :

Solve the following system of equations using cross multiplication method.

2x + 7y - 5  =  0

-3x + 8y  =  -11

Solution:

First we have to change the given linear equations in the form a₁x + b₁y + c₁  =  0, a₂x + b₂y + c₂  =  0.

2x + 7y - 5  =  0

-3x + 8y + 11  =  0

x/(77 + 40)  =  y/(15 - 22)  =  1/[16 + 21]

x/117  =  y/(-7)  =  1/37

So, the solution is

(x, y)  =  (117/37, -7/37)

Example 5 :

Solve the following system of equations by graphing.

x + y - 4  =  0 

3x - y  =  0

Solution :

Step 1 :

Let us re-write the given equations in slope-intercept form  (y = mx + b). 

y  =  - x + 4

(slope is -1 and y-intercept is 4) 

y  =  3x 

(slope is 3 and y-intercept is 0) 

Based on slope and y-intercept, we can graph the given equations. 

Step 2 :

Find the point of intersection of the two lines. It appears to be (1, 3). Substitute to check if it is a solution of both equations.

Because the point (1, 3) satisfies both the equations, the solution for the given system is (1, 3).  

Example 6 :

The school that Jaidee goes to is selling tickets to the annual talent show. On the first day of ticket sales the school sold 10 senior citizen tickets and 11 student tickets for a total of $190. The school took in $160 on the second day by selling 5 senior citizen tickets and 12 student  tickets. Find the price of a senior citizen ticket and the price of a student ticket.

Solution :

Let x be the price of senior citizen ticket and y be the cost of student ticket.

10x + 11y = 190 ----(1)

5x + 12y = 160 ----(2)

(1) - (2) ⋅ 2

10x + 11y - (10x + 24y) = 190 - 320

10x + 11y - 10x - 24y = -130

-13y = -130

y = 10

Applying the value of y, we get

10x + 11(10) = 190

10x + 110 = 190

10x = 190 - 110

10x = 80

x = 80/10

x = 8

So, the cost of 1 senior ticket is $8 and cost of one student ticket is $10.

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