Solve the following pair of linear equations using cross multiplication method :
`{:(2x-5y=14),(x+2y=-2):}`

Answer

To solve the given pair of linear equations using the cross multiplication method, we will follow these steps:

Given equations:
1. \( 2x - 5y = 14 \) (Equation 1)
2. \( x + 2y = -2 \) (Equation 2)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations to set them equal to zero:
- From Equation 1: \( 2x - 5y - 14 = 0 \)
- From Equation 2: \( x + 2y + 2 = 0 \)

Now we have:
1. \( 2x - 5y - 14 = 0 \) (Equation 1)
2. \( x + 2y + 2 = 0 \) (Equation 2)

### Step 2: Identify coefficients
From the equations, we identify the coefficients:
- For Equation 1: \( a_1 = 2, b_1 = -5, c_1 = -14 \)
- For Equation 2: \( a_2 = 1, b_2 = 2, c_2 = 2 \)

### Step 3: Set up the cross multiplication
Using the cross multiplication method, we set up the following:
\[ \frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1} \]

Substituting the coefficients:
\[ \frac{x}{(-5)(2) - (2)(-14)} = \frac{y}{(-14)(1) - (2)(2)} = \frac{1}{(2)(2) - (1)(-5)} \]

### Step 4: Calculate the determinants
Now we calculate each part:
1. For \( x \):
\[ b_1 c_2 - b_2 c_1 = (-5)(2) - (2)(-14) = -10 + 28 = 18 \]
Thus, \( \frac{x}{18} \)

2. For \( y \):
\[ c_1 a_2 - c_2 a_1 = (-14)(1) - (2)(2) = -14 - 4 = -18 \]
Thus, \( \frac{y}{-18} \)

3. For the constant term:
\[ a_1 b_2 - a_2 b_1 = (2)(2) - (1)(-5) = 4 + 5 = 9 \]
Thus, \( \frac{1}{9} \)

### Step 5: Write the equations
Now we have:
\[ \frac{x}{18} = \frac{y}{-18} = \frac{1}{9} \]

### Step 6: Solve for \( x \) and \( y \)
1. From \( \frac{x}{18} = \frac{1}{9} \):
\[ x = 18 \cdot \frac{1}{9} = 2 \]

2. From \( \frac{y}{-18} = \frac{1}{9} \):
\[ y = -18 \cdot \frac{1}{9} = -2 \]

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