## SYSTEM OF LINEAR EQUATIONS IN TWO VARAIBLES

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 = -13x-y  =  73(2)-(-1)  =  76+1  =  77  =  7 x = 2 and y = -12x+3y  =  12(2)+3(-1)  =  14-3  =  11  =  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 :

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 :

Solve the following system of equations by substitution method.

2x - 3y = -1  and  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

-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 :

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 applying the value of x  =  4 in (1), we get

4 + 2y  =  7

2y  =  7 – 4

2 y  =  3

y  =  3/2

Hence the solution is (4, 3/2)

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

Example :

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 :

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

So the value of k is -16.

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