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

a_{1}x+b_{1}y+c_{1} = 0

a_{2}x+b_{2}y+c_{2} = 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 :**

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 :**

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

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 :**

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 :**

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).

**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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