Solve :
`(7 + x)/(5) - (2x - y)/(4) = 3y - 5`
`(5y - 7)/(2) + (4x - 3)/(6) = 18 - 5x`

Answer

To solve the simultaneous equations given by:

1. \(\frac{7 + x}{5} - \frac{2x - y}{4} = 3y - 5\)
2. \(\frac{5y - 7}{2} + \frac{4x - 3}{6} = 18 - 5x\)

we will follow these steps:

### Step 1: Clear the fractions in the first equation

Multiply the entire equation by the least common multiple (LCM) of the denominators, which is 20.

\[ 20 \left(\frac{7 + x}{5}\right) - 20 \left(\frac{2x - y}{4}\right) = 20(3y - 5) \]

This simplifies to:

\[ 4(7 + x) - 5(2x - y) = 60y - 100 \]

### Step 2: Distribute and simplify the first equation

Distributing gives:

\[ 28 + 4x - 10x + 5y = 60y - 100 \]

Combine like terms:

\[ -6x + 5y + 28 = 60y - 100 \]

Rearranging gives:

\[ -6x - 55y = -128 \quad \text{(Equation 1)} \]

### Step 3: Clear the fractions in the second equation

Multiply the entire second equation by the LCM of the denominators, which is 6.

\[ 6 \left(\frac{5y - 7}{2}\right) + 6 \left(\frac{4x - 3}{6}\right) = 6(18 - 5x) \]

This simplifies to:

\[ 3(5y - 7) + (4x - 3) = 108 - 30x \]

### Step 4: Distribute and simplify the second equation

Distributing gives:

\[ 15y - 21 + 4x - 3 = 108 - 30x \]

Combine like terms:

\[ 4x + 15y - 24 = 108 - 30x \]

Rearranging gives:

\[ 34x + 15y = 132 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations

Now we have the two equations:

1. \(-6x - 55y = -128\)
2. \(34x + 15y = 132\)

To eliminate \(x\), we can multiply the first equation by 34 and the second equation by 6:

\[ 34(-6x - 55y) = 34(-128) \implies -204x - 1870y = -4352 \quad \text{(Equation 3)} \]

\[ 6(34x + 15y) = 6(132) \implies 204x + 90y = 792 \quad \text{(Equation 4)} \]

### Step 6: Add the equations

Adding Equation 3 and Equation 4:

\[ (-204x - 1870y) + (204x + 90y) = -4352 + 792 \]

The \(x\) terms cancel out:

\[ -1870y + 90y = -4352 + 792 \]

This simplifies to:

\[ -1780y = -3560 \]

### Step 7: Solve for \(y\)

Dividing both sides by -1780:

\[ y = 2 \]

### Step 8: Substitute \(y\) back to find \(x\)

Substituting \(y = 2\) into Equation 1:

\[ -6x - 55(2) = -128 \]

This simplifies to:

\[ -6x - 110 = -128 \]

Rearranging gives:

\[ -6x = -128 + 110 \]

\[ -6x = -18 \]

Dividing both sides by -6:

\[ x = 3 \]

### Final Solution

Thus, the solution to the simultaneous equations is:

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

---