Use method of cross-multiplications to solve:
` 2x + y = 8 and 3x- 2y = 5`

Answer

To solve the system of equations using the method of cross-multiplication, we start with the given equations:

1. \( 2x + y = 8 \)
2. \( 3x - 2y = 5 \)

### Step 1: Reformat the equations
We need to rewrite both equations in the standard form \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \).

Rearranging the first equation:
\[ 2x + y - 8 = 0 \quad \Rightarrow \quad 2x + 1y - 8 = 0 \]

Rearranging the second equation:
\[ 3x - 2y - 5 = 0 \quad \Rightarrow \quad 3x - 2y - 5 = 0 \]

Now we have:
- \( a_1 = 2, b_1 = 1, c_1 = 8 \)
- \( a_2 = 3, b_2 = -2, c_2 = 5 \)

### Step 2: Apply the cross-multiplication formulas
Using the cross-multiplication method, we can find \( x \) and \( y \) using the formulas:
\[ x = \frac{b_1c_2 - c_1b_2}{a_1b_2 - b_1a_2} \]
\[ y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - b_1a_2} \]

### Step 3: Calculate \( x \)
Substituting the values into the formula for \( x \):
\[ x = \frac{(1)(5) - (8)(-2)}{(2)(-2) - (1)(3)} \]
Calculating the numerator:
\[ 1 \cdot 5 = 5 \]
\[ 8 \cdot -2 = -16 \quad \Rightarrow \quad -(-16) = 16 \]
So, the numerator becomes:
\[ 5 + 16 = 21 \]

Calculating the denominator:
\[ (2)(-2) = -4 \]
\[ (1)(3) = 3 \]
So, the denominator becomes:
\[ -4 - 3 = -7 \]

Now substituting back:
\[ x = \frac{21}{-7} = -3 \]

### Step 4: Calculate \( y \)
Substituting the values into the formula for \( y \):
\[ y = \frac{(8)(3) - (5)(2)}{(2)(-2) - (1)(3)} \]
Calculating the numerator:
\[ 8 \cdot 3 = 24 \]
\[ 5 \cdot 2 = 10 \]
So, the numerator becomes:
\[ 24 - 10 = 14 \]

Using the same denominator as before:
\[ -4 - 3 = -7 \]

Now substituting back:
\[ y = \frac{14}{-7} = -2 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 3, \quad y = 2 \]