`1/16x+1/15y=9/20; 1/20x-1/27y=4/45`

Answer

To solve the given system of equations:

1. **First Equation:**
\[ \frac{1}{16}x + \frac{1}{15}y = \frac{9}{20} \]

2. **Second Equation:**
\[ \frac{1}{20}x - \frac{1}{27}y = \frac{4}{45} \]

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

To eliminate the fractions, we can multiply through by the least common multiple (LCM) of the denominators (16, 15, and 20). The LCM of 16, 15, and 20 is 240.

Multiplying the entire equation by 240 gives:
\[ 240 \left(\frac{1}{16}x\right) + 240 \left(\frac{1}{15}y\right) = 240 \left(\frac{9}{20}\right) \]

This simplifies to:
\[ 15x + 16y = 108 \quad \text{(Equation 1)} \]

### Step 2: Clear the fractions in the second equation

Now, we do the same for the second equation. The LCM of 20 and 27 is 540.

Multiplying the entire equation by 540 gives:
\[ 540 \left(\frac{1}{20}x\right) - 540 \left(\frac{1}{27}y\right) = 540 \left(\frac{4}{45}\right) \]

This simplifies to:
\[ 27x - 20y = 48 \quad \text{(Equation 2)} \]

### Step 3: Solve for y in terms of x from Equation 1

From Equation 1:
\[ 15x + 16y = 108 \]

Rearranging gives:
\[ 16y = 108 - 15x \]
\[ y = \frac{108 - 15x}{16} \quad \text{(Equation 3)} \]

### Step 4: Substitute Equation 3 into Equation 2

Substituting Equation 3 into Equation 2:
\[ 27x - 20\left(\frac{108 - 15x}{16}\right) = 48 \]

Multiplying through by 16 to eliminate the fraction:
\[ 16 \cdot 27x - 20(108 - 15x) = 48 \cdot 16 \]
\[ 432x - 2160 + 300x = 768 \]

### Step 5: Combine like terms and solve for x

Combining like terms:
\[ 732x - 2160 = 768 \]

Adding 2160 to both sides:
\[ 732x = 768 + 2160 \]
\[ 732x = 2928 \]

Dividing by 732:
\[ x = \frac{2928}{732} = 4 \]

### Step 6: Substitute x back to find y

Now, substitute \(x = 4\) back into Equation 3:
\[ y = \frac{108 - 15(4)}{16} \]
\[ y = \frac{108 - 60}{16} \]
\[ y = \frac{48}{16} = 3 \]

### Final Solution

The solution to the system of equations is:
\[ x = 4, \quad y = 3 \]