# Solve the following systems of equations for x and y: 8x+3y=1 and 6x+2y=4.

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You also may use the following alternative method, hence, you should evaluate the determinant of matrix of coefficients of varaibles x and y, such that:

`Delta = [(8,3),(6,2)]` => `Delta = 16 - 18 => Delta = -2`

Since `Delta != 0` , hence, you may use Cramer's method to evaluate variables `x, y` , such that:

`x = Delta_x/Delta`

`Delta_x = [(1,3),(4,2)]` => `Delta_x = 2 - 12 => Delta_x = -10`

`x = (-10)/(-2) => x = 5`

`y = Delta_y/Delta`

`Delta_y = [(8,1),(6,4)] => ` `Delta_y = 32 - 6 => Delta_y = 26`

`y = 26/(-2) => y = -13`

**Hence, evaluating the solution to the system, using Cramer's method, yields **`x = 5, y = -13.`

The system of equations 8x+3y=1 and 6x+2y=4 has to be solved.

From 6x + 2y = 4 we get

3x + y = 2

y = 2 - 3x

Substitute this for y in 8x+3y=1

8x + 3(2 - 3x) = 1

8x + 6 - 9x = 1

-x = -5

x = 5

y = 2 - 3*x = 2 - 15 = -13

The solution of the given system of equations is x = 5 and y = -13

We'll solve the system using elimination method.

We notice that the second equation of the system could be divided by 2:

6x + 2y = 4

3x + y = 2 (2)

8x + 3y = 1 (1)

We'll multiply by -3 the equation (2):

-9x -3y = -6 (3)

We'll add (3)+(1):

-9x -3y + 8x + 3y = -6 + 1

We'll combine and eliminate like terms:

-x = -5

x = 5

We'll substitute x in (1):

40 + 3y = 1

3y = 1 - 40

3y = -39

y = -13

The solution of the system is: {5 , -13}.