You may use matrices to solve the system, hence, you need to create the matrix of coefficients of variables, such that:

A = ((3,-8),(2,3))

You need to evaluate the determinant of the matrix A, such that:

`det A = [(3,-8),(2,3)]` => `det A = 9 + 16 => detA = 25`

You need to use Cramer's formulas to evaluate x and y, such that:

`x = (Delta_x)/(det A); y = (Delta_y)/(det A);`

`Delta_x = [(10,-8),(6,3)]` => `Delta_x =30 + 48 = 78`

`x = 78/25`

`Delta_y = [(3,10),(2,6)]` =>` Delta_y = 18 - 20 = -2`

`y = -2/25`

**Hence, evaluating the solution to the system of equations, using Cramer's formulas, yields `x = 78/25, y = -2/25` .**

The system of equations consisting of 3x - 8y = 10 and 2x + 3y = 6 has to be solved for x and y.

3x - 8y = 10

=> 3x = 10 + 8y

=> x = 10/3 + (8y)/3

Substitute this for x in 2x + 3y = 6

=> 2*(10/3 + (8y)/3) + 3y = 6

=> 20/3 + (16*y)/3 + 3y = 6

=> (25y)/3 = 6 - 20/3

=> 25y = -2

=> y = -2/25

x = 10/3 + (8y)/3 = 10/3 - 16/75 = 78/25

**The solution of the given set of equations is x = 78/25 and y = -2/25**

3x - 8y = 10 ------eq(i)

2x + 3y = 6 ------eq(ii)

To find the value of x and y we can use the substitution method by considering eq (i) to isolate y:

3x - 8y = 10

-8y=10-3x

y=(10-3x)/-8

y=-(-10+3x)/-8

On cancelling off the minus signs we get:

y= (-10+3x)/8

y= -10/8 + 3x/8

Putting this value in eq(ii):

2x + 3{-10/8 + 3x/8} = 6

Multiplying 3 with each component in the bracket:

2x -30/8 + 9x/8 = 6

Taking 8 as L.C.M.:

{16x -30 +9x}/8 = 6

Taking 8 on right hand side:

16x - 30 +9x = 6*8

16x - 30 +9x = 48

25x - 30 = 48

25x = 48 +30

25x = 78

**x=78/25**

Putting value of x in eq(i) to find out y:

3x - 8y = 10

3(78/25) - 8y = 10

234/25 - 8y = 10

taking 234/25 on the right hand side:

-8y = 10 - 234/25

Taking 25 as L.C.M:

-8y = {250-234}/25

-8y = 16/25

Taking -8 on right hand side:

y = -16/25*-8

y = -16/200

**y = -2/25**

**Therefore: x=78/25 and y = -2/25**