Solve, using cross - multiplication :
3x + 4y = 11
2x + 3y = 8

Answer

To solve the system of equations using cross multiplication, we have the equations:

1. \(3x + 4y = 11\) (Equation 1)
2. \(2x + 3y = 8\) (Equation 2)

### Step 1: Rewrite the equations in standard form
We can rewrite the equations in the form \(Ax + By + C = 0\):

1. \(3x + 4y - 11 = 0\) (Here, \(a_1 = 3\), \(b_1 = 4\), \(c_1 = -11\))
2. \(2x + 3y - 8 = 0\) (Here, \(a_2 = 2\), \(b_2 = 3\), \(c_2 = -8\))

### Step 2: Identify coefficients
From the equations, we have:
- \(a_1 = 3\), \(b_1 = 4\), \(c_1 = -11\)
- \(a_2 = 2\), \(b_2 = 3\), \(c_2 = -8\)

### Step 3: Set up the cross multiplication formula
The cross multiplication formula is given by:

\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]

### Step 4: Calculate the determinants
1. For \(b_1c_2 - b_2c_1\):
\[ b_1c_2 - b_2c_1 = 4 \cdot (-8) - 3 \cdot (-11) = -32 + 33 = 1 \]

2. For \(c_1a_2 - c_2a_1\):
\[ c_1a_2 - c_2a_1 = (-11) \cdot 2 - (-8) \cdot 3 = -22 + 24 = 2 \]

3. For \(a_1b_2 - a_2b_1\):
\[ a_1b_2 - a_2b_1 = 3 \cdot 3 - 2 \cdot 4 = 9 - 8 = 1 \]

### Step 5: Substitute into the cross multiplication formula
Now we can substitute these values into the formula:

\[ \frac{x}{1} = \frac{y}{2} = \frac{1}{1} \]

### Step 6: Solve for \(x\) and \(y\)
From the above equations, we have:
1. \(x = 1\)
2. \(\frac{y}{2} = 1 \implies y = 2\)

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 1, \quad y = 2 \]