SYSTEM OF LINEAR EQUATIONS IN TWO VARIABLES

A linear system of two equations with two variables is any system that can be written in the form.

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

A solution to a system of equations is a value of x and and value of y, when substituted into the equations, satisfies both equations at the same time.

For the example,

3x - y = 7 ----(1)

2x + 3y = 1 ----(2)

3(1) + (2)

9x - 3y + 2x + 3y = 21 + 2

11x = 22

x = 2

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

3(2) - y = 7

6 - y = 7

y = -1

x = 2 and y = −1 is a solution to the system.

Checking :

3x - y = 7 ----(1)

2x + 3y = 1 ----(2)

x = 2 and y = -1

3x - y = 7

3(2) - (-1) = 7

6 + 1 = 7

7 = 7

x = 2 and y = -1

2x + 3y = 1

2(2) + 3(-1) = 1

4 - 3 = 1

1 = 1

The following method are used to solve system of linear equations

(i)  Graphing

(ii)  Substitution 

(iii)  Elimination

(iv)  Cross multiplication

Example Problem of Solving System of Linear Equations By Graphing Method

Example 1 :

x + y = 7

x - y = 3

Solution :

x + y = 7 ----(1)

x - y = 3 ----(2)

From (1),

y = 7 - x

Substitute some random values for x and solve for y. 

The points on the first line are

(-1, 8), (0, 7), (1, 6)

From (2),

y = x - 3

Substitute some random values for x and solve for y. 

The points on the second line are

(-1, -4), (0, -3), (1, -2)

Plot the points from the above tables and sketch the graph

The point of intersection of the two lines is the solution.

So, the solution is (x, y) = (5, 2).

Example Problem of Solving System of Linear Equations By Substitution Method

Example 2 :

Solve the following system of equations by substitution method.

2x - 3y = -1

y = x - 1

Solution :

2x - 3y = -1 ----(1) 

y = x - 1 ----(2) 

Step 1 :

From (2), substitute (x - 1) for y into (1).

(1)----> 2x - 3(x - 1) = -1

Simplify.

2x - 3x + 3 = -1

-x - 3 = -1

Add 3 to each side. 

-x = 2

Multiply each side (-1). 

x = -2

Step 2 : 

Substitute -2 for x into (2). 

(2)----> y = 2 - 1

 y = 1

Therefore, the solution is 

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

Example Problem of Solving System of Linear Equations By Elimination Method

Example 3 :

Solve the following system of linear equations by elimination method.

x + 2y = 7

x – 2y = 1 

Solution :

x + 2y = 7 ----(1)

x – 2y = 1 ----(2)

The coefficients of x and y are equal in both the equations.

(1) + (2) :

2x = 8

x = 8/2

x = 4

By substituting x = 4 in (1), we get

4 + 2y = 7

2y = 7 – 4

2y = 3

y = 3/2

Hence the solution is (4, 3/2).

Example Problem of Solving System of Linear Equations By Cross Multiplication Method

Example 4 :

Solve the following system of linear equations by cross multiplication method.

3x + 4y = 24

20x - 11y = 47

Solution :

3x + 4y - 24 = 0 ----(1)

20x - 11y - 47 = 0 ----(2)

x/(-188 - 264) = y/(-480 -(-141)) = 1/(-33 - 80)

x/(-452) = y/(-480 + 141)) = 1/(-33 - 80)

x/(-452) = y/(-339) = 1/(-113)

x/(-452) = 1/(-113)

x = 4

x/(-339) = 1/(-113)

x = 3

So, the solution is (4, 3).

Finding Unknown When Solution of Linear Equation is Given

Example 5 :

If the point (4,-2) lies on the line 5x + 2y + k = 0.

Find the value of k.

Solution :

Here the given point (4,-2) lies on the given line.

To solve this problem we have to apply the given points in the equation as the value of x and y.

Here x = 4 and y = -2

The given equation is 5x + 2y + k = 0.

By applying the values of x and y, we get

5(4) + 2(-2) + k = 0

20 - 4 + k = 0

16 + k = 0

k = -16

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