If `x/6+y/15=4` and `x/3-y/12=4 3/4` find x and y.

Options

A. (6,15)
B. (18,15)
C. (15,18)
D. (12,30)

Answer

To solve the equations \( \frac{x}{6} + \frac{y}{15} = 4 \) and \( \frac{x}{3} - \frac{y}{12} = 4 \frac{3}{4} \), we will follow these steps:

### Step 1: Rewrite the equations
The first equation is:
\[ \frac{x}{6} + \frac{y}{15} = 4 \]
To eliminate the fractions, we can multiply the entire equation by 30 (the least common multiple of 6 and 15):
\[ 30 \left(\frac{x}{6}\right) + 30 \left(\frac{y}{15}\right) = 30 \cdot 4 \]
This simplifies to:
\[ 5x + 2y = 120 \quad \text{(Equation 1)} \]

The second equation is:
\[ \frac{x}{3} - \frac{y}{12} = 4 \frac{3}{4} \]
First, convert \( 4 \frac{3}{4} \) to an improper fraction:
\[ 4 \frac{3}{4} = \frac{19}{4} \]
Now, multiply the entire equation by 12 (the least common multiple of 3 and 12):
\[ 12 \left(\frac{x}{3}\right) - 12 \left(\frac{y}{12}\right) = 12 \cdot \frac{19}{4} \]
This simplifies to:
\[ 4x - y = 57 \quad \text{(Equation 2)} \]

### Step 2: Solve the system of equations
Now we have the two equations:
1. \( 5x + 2y = 120 \)
2. \( 4x - y = 57 \)

To eliminate \( y \), we can multiply Equation 2 by 2:
\[ 2(4x - y) = 2(57) \]
This gives us:
\[ 8x - 2y = 114 \quad \text{(Equation 3)} \]

### Step 3: Add Equation 1 and Equation 3
Now we can add Equation 1 and Equation 3:
\[ (5x + 2y) + (8x - 2y) = 120 + 114 \]
This simplifies to:
\[ 13x = 234 \]
Now, solve for \( x \):
\[ x = \frac{234}{13} = 18 \]

### Step 4: Substitute \( x \) back to find \( y \)
Now substitute \( x = 18 \) back into Equation 1:
\[ 5(18) + 2y = 120 \]
This simplifies to:
\[ 90 + 2y = 120 \]
Now, isolate \( 2y \):
\[ 2y = 120 - 90 \]
\[ 2y = 30 \]
Now, solve for \( y \):
\[ y = \frac{30}{2} = 15 \]

### Final Answer
Thus, the solution is:
\[ x = 18, \quad y = 15 \]