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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