Solve, using cross - multiplication :
4x - 3y = 0
2x + 3y = 18

Answer

To solve the simultaneous equations using cross-multiplication, we have the following equations:

1. \(4x - 3y = 0\)
2. \(2x + 3y = 18\)

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

From the first equation:
- \(a_1 = 4\)
- \(b_1 = -3\)
- \(c_1 = 0\)

From the second equation:
- \(a_2 = 2\)
- \(b_2 = 3\)
- \(c_2 = -18\)

### Step 2: Use the cross-multiplication formula
The formulas for \(x\) and \(y\) using cross-multiplication are:

\[ x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1} \]
\[ y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1} \]

### Step 3: Substitute the values into the formulas
Substituting the values we found into the formulas:

For \(x\):
\[ x = \frac{(-3)(-18) - (3)(0)}{(4)(3) - (2)(-3)} \]

Calculating the numerator:
\[ = \frac{54 - 0}{12 + 6} \]

Calculating the denominator:
\[ = \frac{54}{18} \]

So,
\[ x = 3 \]

For \(y\):
\[ y = \frac{(0)(2) - (-18)(4)}{(4)(3) - (2)(-3)} \]

Calculating the numerator:
\[ = \frac{0 + 72}{12 + 6} \]

So,
\[ y = \frac{72}{18} \]

Thus,
\[ y = 4 \]

### Final Solution
The solution to the equations is:
\[ x = 3, \quad y = 4 \]

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