To solve system of linear equations in two variable, we use the following rules.

**Rule 1 :**

If ∆ ≠ 0. Then the system has unique solution and we can solve the equations by using the formula

x = ∆ₓ/∆ , y = ∆ᵧ/∆

**Rule 2 :**

If

∆ = 0 and ∆ₓ = 0, ∆ᵧ = 0

and at least one of the coefficients a_{11}, a_{12}, a_{21}, a_{22} is non zero, then the system is consistent and has infinitely many solution.

**Rule 3 :**

If ∆ = 0 and at least one of the values ∆ₓ, ∆ᵧ is non-zero then the system is inconsistent and it has no solution.

**Example 1 :**

Solve the following equation using determinant method

x + 2y = 3, x + y = 2

**Solution :**

Write the values of Δ, Δx and Δy and evaluate

Here Δ ≠ 0, Δx ≠ 0 and Δy ≠ 0.

So, the system is consistent and it has unique solution.

x = Δx/Δ and y = Δy/Δ

x = -1/(-1) ==> 1

x = -1/(-1) ==> 1

Hence the solution is (1, 1).

**Example 2 :**

Solve the following equation using determinant method

3x + 2y = 5 and x + 3y = 4

**Solution :**

Here Δ ≠ 0, Δx ≠ 0 and Δy ≠ 0.

So, the system is consistent and it has unique solution.

x = Δx/Δ and y = Δy/Δ

x = 7/7 ==> 1

x = 7/7 ==> 1

Hence the solution is (1, 1).

**Example 3 :**

Solve the following equation using determinant method

x + 2y = 3 and 2x + 4y = 8

**Solution :**

Here, Δ = 0 but Δx ≠ 0.

So, the system is inconsistent and it has no solution.

**Example 4 :**

Solve the following equation using determinant method

x + 2y = 3 and 2x + 4y = 6

**Solution :**

Since ∆ = 0, ∆**ₓ** = 0 and ∆**ᵧ** = 0 and atleast one of the element in ∆ is non zero.

Then the system is consistent and it has infinitely many solution. The above system is reduced into one equation. To solve this equation we have to assign y = k.

x+2y = 3

x+2(k) = 3

x+2k = 3

x = 3-2k and y = k

So, the solution is (3-2k, k). Here k ∈ R where R is real numbers.

**Example 5 :**

Solve the following equation using determinant method

2x+4y = 6, 6x+12y = 24

**Solution :**

Here ∆ = 0 but ∆**ₓ** ≠ 0, then the system is consistent and it has no solution.

**Example 6 :**

Solve the following equation using determinant method

2x+y = 3 and 6x+3y = 9

**Solution :**

Since ∆ = 0, ∆**ₓ** = 0 and ∆**ᵧ** = 0 and atleast one of the element in ∆ is non zero. Then the system is consistent and it has infinitely many solution. The above system is reduced into single equation. To solve this equation we have to assign y = k.

2x+y = 3

2x+k = 3

2x+k = 3

2x = 3-k

x = (3-k)/2

y = k

So, the solution is ((3-k)/2, k). Here k ∈ R where R is real numbers.

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