**Cross Multiplication Method :**

This is the method which can be used to solve linear equations.

Let us consider the following system of linear equations.

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

a_{2}x + b_{2}y + c_{2} = 0

We have to write the coefficients of the equations and do cross multiplication as shown below.

We write the coefficient of y and constant term and two more columns by repeating the coefficients of x and y as follows.

The result is given by

The solution is

**Example 1 :**

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.

That is we have to bring the constants with x, y terms.

2 x + 7 y - 5 = 0

-3 x + 8 y + 11 = 0

x/[(7(11) - (8) (-5)] = y/[(-5(-3) - (2) (11)] = 1/[(2(8) - (-3) (7)]

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

x/[117] = y/[-7] = 1/[37]

x/[117] = 1/[37] x = 117/37 |
y/[-7] = 1/[37] y = -7/37 |

Therefore,the solution is (117/37,-7/37)

**Example 2 :**

**Solve the following system of equations using cross multiplication method :**

3x + 4y = 24

20x - 11y = 47

**Solution:**

First we have to make the given equations in the form of a₁ x + b₁ y + c₁ = 0,a₂ x + b₂ y + c₂ = 0.

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

20 x - 11 y - 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 = (-452)/(-113) x = 4 |
y/(-339) = 1/(-113) y = (-339)/(-113) y = 3 |

Therefore solution is (4,3).

**Verification:**

Now let us apply the answer that we got in the first or second equation to check whether we got correct answer or not.

3 x + 4 y = 24

3(4) + 4(3) = 24

12 + 12 = 24

24 = 24

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