Solve, using cross - multiplication :
`4x - 7y + 28 = 0 and -7x + 5y + 9 = 0`.

Answer

To solve the equations \(4x - 7y + 28 = 0\) and \(-7x + 5y + 9 = 0\) using cross-multiplication, we will follow these steps:

### Step 1: Rewrite the equations in standard form
The given equations are:
1. \(4x - 7y + 28 = 0\) can be rewritten as \(4x - 7y = -28\).
2. \(-7x + 5y + 9 = 0\) can be rewritten as \(-7x + 5y = -9\).

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \(4x - 7y = -28\):
- \(a_1 = 4\)
- \(b_1 = -7\)
- \(c_1 = -28\)

- For the second equation \(-7x + 5y = -9\):
- \(a_2 = -7\)
- \(b_2 = 5\)
- \(c_2 = -9\)

### Step 3: Set up the cross-multiplication formula
The cross-multiplication formula is given by:
\[ \frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1} \]

### Step 4: Substitute the values into the formula
Now, substituting the values into the formula:
\[ \frac{x}{(-7)(-9) - (5)(-28)} = \frac{y}{(-28)(-7) - (-9)(4)} = \frac{1}{(4)(5) - (-7)(-7)} \]

### Step 5: Calculate the determinants
1. For \(b_1 c_2 - b_2 c_1\):
\[ (-7)(-9) - (5)(-28) = 63 + 140 = 203 \]

2. For \(c_1 a_2 - c_2 a_1\):
\[ (-28)(-7) - (-9)(4) = 196 + 36 = 232 \]

3. For \(a_1 b_2 - a_2 b_1\):
\[ (4)(5) - (-7)(-7) = 20 - 49 = -29 \]

### Step 6: Write the equations from the results
Now we have:
\[ \frac{x}{203} = \frac{y}{232} = \frac{1}{-29} \]

### Step 7: Solve for \(x\) and \(y\)
From \(\frac{x}{203} = \frac{1}{-29}\):
\[ x = \frac{203}{-29} = -7 \]

From \(\frac{y}{232} = \frac{1}{-29}\):
\[ y = \frac{232}{-29} = -8 \]

### Final Solution
Thus, the solution to the equations is:
\[ x = 7, \quad y = 8 \]