CRAMER RULE FOR 2 EQUATIONS

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

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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