Solve, using cross - multiplication :
4x + 3y = 17
3x - 4y + 6 =0

Answer

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

1. \( 4x + 3y = 17 \)
2. \( 3x - 4y + 6 = 0 \)

First, we need to rewrite the second equation in the standard form \( a_2x + b_2y + c_2 = 0 \):

\[ 3x - 4y + 6 = 0 \implies 3x - 4y = -6 \]

Now we can express both equations in the standard form:

1. \( 4x + 3y - 17 = 0 \) (Here, \( a_1 = 4, b_1 = 3, c_1 = -17 \))
2. \( 3x - 4y + 6 = 0 \) (Here, \( a_2 = 3, b_2 = -4, c_2 = 6 \))

Next, we will use the cross-multiplication formula to find \( x \) and \( y \):

The formulas for \( x \) and \( y \) are given by:

\[ x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1} \]

\[ y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \]

Now we will substitute the values into these formulas.

### Step 1: Calculate \( x \)

Substituting the values into the formula for \( x \):

\[ x = \frac{(3)(6) - (-4)(-17)}{(4)(-4) - (3)(3)} \]

Calculating the numerator:

\[ 3 \times 6 = 18 \]
\[ -4 \times -17 = 68 \]

So, the numerator becomes:

\[ 18 - 68 = -50 \]

Now calculating the denominator:

\[ 4 \times -4 = -16 \]
\[ 3 \times 3 = 9 \]

So, the denominator becomes:

\[ -16 - 9 = -25 \]

Now substituting back into the equation for \( x \):

\[ x = \frac{-50}{-25} = 2 \]

### Step 2: Calculate \( y \)

Now we will substitute the values into the formula for \( y \):

\[ y = \frac{(-17)(3) - (6)(4)}{(4)(-4) - (3)(3)} \]

Calculating the numerator:

\[ -17 \times 3 = -51 \]
\[ 6 \times 4 = 24 \]

So, the numerator becomes:

\[ -51 - 24 = -75 \]

Now using the same denominator as before, which is \(-25\):

Now substituting back into the equation for \( y \):

\[ y = \frac{-75}{-25} = 3 \]

### Final Solution

Thus, the solution to the simultaneous equations is:

\[ x = 2, \quad y = 3 \]